ELEMENTARY CALCULUS

BY

FREDERICK S. WOODSAND

FREDERICK H. BAILEY

PIlOlfMSSOUS 0V MATHEMATICS IN THK MASSACHUSETTS

INSTITUTE OF TECHNOLOGY

GINN AND COMPANYItORVOK NI3W YORK OHtCAGO LONDON

ATLANTA AM.A8 COLUMBUS SAN PBANOISOO

COPYRIGHT, 1922, BY FREDERICK 8 WOODSAND FREDERICK H BAILEY

ALL BIGHTS RESERVED

PRINTED IN THB UNITED STATES OF AMERICAI

82611

153

GINN AND COMPANY . PRO-PRIETORS BOSTON > V.S.A.

PKEFACB

This book is adapted to the use of students in the first year

in technical school or college, and is based upon the experience

of the authors m teaching calculus to students in the Massa-

chusetts Institute of Technology immediately upon entrance.

It is accordingly assumed that the student has had college-

entrance algebra, including graphs, and an elementary course

in trigonometry, but that he has not studied analytic geometry.The first three chapters form an introductory course in

which the fundamental ideas of the calculus are introduced,

including derivative, differential, and the definite integral, but

the formal work is restricted to that involving only the poly-nomial. These chapters alone are well fitted for a short course

of about a term.

The definition of the derivative is obtained through the

concept of speed, using familiar illustrations, and the idea

of a derivative as measuring the rate of change of related

quantities is emphasized. The slope of a curve is introduced

later. This is designed to prevent the student from acquiringthe notion that the derivative is fundamentally a geometric

concept. For the same reason, problems from mechanics are

prominent throughout the book.

With Chapter IV a more formal development of the subject

begins, and certain portions of analytic geometry are introducedas needed. These include, among other things, the straight line,

the conic sections, the cycloid, and polar coordinates.

The book contains a large number of well-graded exercises forthe student. Drill exercises are placed at the end of most sec-

tions, and a miscellaneous 'Set of exercises, for review or further

work, is found at the end of each chapter except the first.

ui

iv PREFACE

Throughout the book, the authors believe, the matter is pre-

sented in a manner which is well within the capacity of a first-

year student to understand. They have endeavored to teach

the calculus from a common-sense standpoint as a very useful

tool. They have used as much mathematical rigor as the

student is able to understand, but have refrained from raising

the more difficult questions which the student in his first

course is able neither to appreciate nor to master.

Students who have completed this text and wish to continue

their study of mathematics may next take a brief course in

differential equations and then a course in advanced calculus,

or they may take a course m advanced calculus which includes

differential equations. It would also be desirable for such stu-

dents to have a brief course in analytic geometry, which mayeither follow this text directly or come later.

This arrangement of work the authors consider preferable to

the one for a long time common in American colleges by

which courses in higher algebra and analytic geometry precede

the calculus. However, the teacher who prefers to follow the

older arrangement will find this text adapted to such a program.

F. S. WOODSF H BAILEY

CONTENTS

CHAPTER I. EATESSECTION PAGE1 Limits . 1

2 Average speed . . . .... 3

3. True speed ... . 5

4. Algebiaic method . .... . 8

5 Acceleiation . ....... 9

G Rate of change ........ .11

CHAPTER II. DIFFERENTIATION

7 The derivative . . . . . .... .158. Differentiation of a polynomial ... 18

9 Sign of the derivative ... . . . .... 20

10 Velocity and acceleration (continued) . ... 21

11 Rate of change (continued) . ... ... 24

12 Graphs . .... .... 27

13 Real roots of an equation 30

14 Slope of a straight line . .... .3115 Slope of a curve ... . 36

16 The second derivative . . , .3917. Maxima and minima . . ... 41

18 Integration ... ... 44

19 Area . 47

20 Differentials ...*... 50

21. Appioximations ... . ... 53

General exercises .... . . . . - . .55

CHAPTER III. SUMMATION

22. Area by summation . . . ... 00

23. The definite integral .... . 62

24. The general summation problem .6625. Pressure 68

26 Yolume . 71

General exercises ... 76

T

vi CONTENTS

CHAPTER IV. ALGEBRAIC FUNCTIONSSECTION PAGE

27 Distance between two points ... . . . . 79

28. Circle ... . .... ... 79

29. Parabola ... 81

30 Parabolic segment . 83

81. Ellipse ... 85

32 Hyperbola . . 87

33. Other curves 91

34 Theorems on limits . . .9335. Theorems on derivatives .... 94

36. Formulas 101

37 Differentiation of implicit functions ... 102

38 Tangent line ... . 104

39. The differentials dx, dy, ds 106

40. Motion in a curve ... ... 107

41. Related velocities and rates Ill

General exercises . 113

CHAPTER V. TRIGONOMETRIC FUNCTIONS

42. Circular measure . .... ... 119

43 Graphs of trigonometric functions . . ... 121

44. Differentiation of trigonometric functions 124

45. Simple harmonic motion . .... 127

46 Graphs of inverse trigonometric functions 130

47 Differentiation of inverse trigonometric functions . . 131

48. Angular velocity . . ... . ... . 185

49. The cycloid . . .... ... . 137

50. Curvature 189

51 Polar coSrdinates,

142

53. The differentials dr, dO, ds, in polar coordinates . . . 146

General exercises . . . . 149

CHAPTER VI. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

53. The exponential function.... 154

54. The logarithm . . 154

55. Certain empirical equations 159

56. Differentiation . . 163

57. The compound-interest law 166

General exercises 168

CONTENTS vii

CHAPTER VII. SERIESSECTION PAGE!

58. Power series . . . . . 172

59. Maclaurm's series . 173

60. Taylor's seuos . . . 177

General exercises . . ... 179

CHAPTER VIII. PARTIAL DIFFERENTIATION

61 Partial differentiation . . . 181

62. Higher partial denvatives . 184

63. Total differential of a function of two variables . . . 185

64 Rate of change ... .189General exercises . . . 191

CHAPTER IX. INTEGRATION

65. Introduction . . . 194

66. Integral of un . . 195

67-68 Other algebraic integrands 199

69 Integrals of trigonometric functions ... 205

70. Integrals of exponential functions . . ... 207

71-72. Substitutions . . . 208

73-74 Integration by parts . . . 212

75 Integration of rational fractions ... . 216

76. Table of integrals . . 217

General exercises ... .... . 220

CHAPTER X. APPLICATIONS

77. Review problems . ... 225

78. Infinite limits, or integrand 229

79. Area in polar cobrdmates . 230

80. Mean value of a function . .... 233

81. Length of a plane curve 235

82. Work 237

General exercises ... . . ... 239

CHAPTER XI. REPEATED INTEGRATION

83. Double integrals ... . . . . 244

84. Area as a double integral . . . . . . 246

85. Center of gravity . 249

yiii CONTENTS

SECTION PAGE

86 Center of gravity of a composite area . . . 255

87 Theoiems . . . 257

88. Moment of inertia 260

89 Moments of inertia about parallel axes . 266

90 Space cooidinates 269

91. Certain suifaces . .... 271

92. Volume . . . .... .27793 Center of gravity of a solid . 282

94 Moment of inertia of a solid . . 283

General exeicises . . . . . 286

ANSWERS 291

INDEX 315

ELEMENTARY CALCULUS

* CHAPTER I

RATES

1. Limits. Since the calculus is based upon the idea of a

limit it is necessary to have a clear understanding of the word.

Two examples already familiar to the student will be sufficient.

In finding the aiea of a circle m plane geometry it is usual

to begin by inscribing a regular polygon in the circle. The area

of the polygon differs from that of the circle by a certain

amount. As the number of sides of the polygon is increased,

this difference becomes less and less. Moreover, if we take any

small number 0, we can find an inscribed polygon whose area

differs from that of the circle by less than e\ and if one such

polygon has been found, any polygon with a larger number of

sides will still differ m area from the circle by less than e. The

area of the circle is said to be the limit of the area of the

inscribed polygon.

As another example of a limit consider the geometric progres-

sion with an unlimited number -of terms

The sum of the first two terms of this series is 1, the sum

of the first three terms is If, the sum of the first four terms

is1-J,

and so on. It may be found by trial and is proved in

the algebras that the sum of the terms becomes more nearly

equal to 2 as the number of terms which are taken becomes

greater. Moreover, it may be shown that if any small number

e is assumed, it is possible to take a number of terms n so that

the sum of these terms differs from 2 by less than e. If a value

of n has thus been found, then the sum of a number of terms

1

2 RATES

greater than n will still differ from 2 by less than e. Thenumber 2 is said to be the limit of the sum of the first n terms

of the series.

In each of these two examples there is a certain variable

namely, the area of the inscribed polygon of n sides in one case

and the sum of the first n terms of the series in the other case

and a certain constant, the area of the circle and the number 2

respectively. In each case the difference between the constant

and the variable may be made less than any small number e by

taking n sufficiently large, and this difference then continues

to be less than e for any larger value of n.

This is the essential property of a limit, which may be defined

as follows:

A constant A is said to be the limit of a variable Xif, as the vari-

able changes its value according to some law, the difference between

the variable and the constant becomes and remains less than anysmall quantity which may be assigned.

The definition does not say that the variable never reaches its

limit. In most cases in this book, however, the variable fails to

do so, as in the two examples already given. For the polygon is

never exactly a circle, nor is the sum. of the terms of the series

exactly 2. Examples may be given, however, of a variable's

becoming equal to its limit, as in the case of a swinging pendulum

finally coming to rest. But the fact that a variable may never

reach its limit does not make the limit inexact. There is nothinginexact about the area of a circle or about the number 2.

The student should notice the significance of the word"remains" in the definition. If a railroad train approaches a

station, the difference between the position of the train and

a point on the track opposite the station becomes less than anynumber which may be named

; but if the train keeps on by the

station, that difference does not remain small. Hence there is

no limit approached in this case.

If X is a variable and A a constant which X approaches as a

limit, it follows from the definition that we may write

X**A + e, (1)

SPEED 3

where e is a quantity (not necessarily positive) which may be

made, and then will remain, as small as we please.

Conversely, if as the result of any reasoning we arrive at a

formula of the form (1) where X is a variable and A a constant,

and if we see that we can make e as small as we please and

that it will then remain just as small or smaller as X varies,

we can say that A is the limit of X. It is in this way that weshall determine limits in the following pages.

2. Average speed. Let us suppose a body (for example, an

automobile) moving from a point A to a point B (Fig. 1), a

distance of 100 mi If the automobile takes 5 hr. for the trip,

we are accustomed to say that it has traveled at the rate of

20 mi. an hour. Everybody knows A p Q Bthat this does not mean that the ' ' '

automobile went exactly 20 mi. in

each hour of the trip, exactly 10 mi. in each half hour, exactly

5 mi in each quarter hour, and so on. Probably no automobile

ever ran in such a way as that. The expression"20 mi. an

hour "may be understood as meaning that a fictitious automobile

traveling in the steady manner* just described would actually

cover the 100 mi. in just 5 hr. ; but for the actual automobile

which made the trip," 20 mi. an hour

"gives only a certain

average speed.

So if a man walks 9 mi. in 3 hr., he has an average speed of

3 mi. an hour. If a stone falls 144 ft. in 3 sec., it has an average

speed of 48 ft. per second. In neither of these cases, however,

does the average speed give us any information as to the actual

speed of the moving object at a given instant of its motion.

The point we are making is so important, and it is so often

overlooked, that we repeat it in the following statement:

If a "body traverses a distance in a certain time, the average speed

of the body in that time is given ly the formula,

, distanceaverage speed =

time

but this formula does not in general give the true speed at any

given time.

4 KATES

EXERCISES

1. A man runs a half mile m 2 mm and 3 sec. What is his

average speed m feet per second ?

2. A man walks a mile m 25 min. What is his average speed in

yards per second ?

3. A train 600 ft. long takes 10 sec. to pass a given milepost.

What is its average speed in miles per hour ?

4. A stone is thrown directly downward from the edge of a

vertical cliff Two seconds afterwards it passes a point 84 ft down

the side of the cliff, and 4 sec after it is thrown it passes a point

296 ft. down the side of the cliff What is the average speed of the

stone in falling between the two mentioned points ?

5. A railroad train runs on the following schedule :

Find the average speed between each two consecutive stations and

for the entire trip.

6. A body moves four times around a circle of diameter 6 ft in

1 min. What is its average speed in feet per second ?

7. A block slides from the top to the bottom of an inclined

plane which makes an angle of 30 with the horizontal. If the topis 50 ft. higher than the bottom and it requires f mm. for the block

to slide down, what is its average speed in feet per second ?

8. Two roads intersect at a point C B starts along one road

toward C from a point 5 mi. distant from C and walks at an average

speed of 3 mi an hour. Twenty minutes later A starts along the

other road toward C from a point 2 mi. away from G At what

average speed must A walk if he is to reach C at the same instant

that B arrives ?

9. A man rows across a river $ mi. wide and lands at a point

mi. farther down the river. If the banks of the river are parallel

straight lines and he takes ^ hr. to cross, what is his average speedin feet per minute if his course is a straight line ?

SPEED 5

10. A trolley car is running along a straight street at an average

speed of 12 mi. per hour. A house is 50 yd. back from the car track

and 100 yd. up the street from a car station. A man comes out of

the house when a car is 200 yd away from the station What must

be the average speed of the man m yards per minute if he goes in

a straight line to the station and arrives at the same instant as

the car ?

3. True speed. How then shall we determine the speed at

which a moving body passes any given fixed point P in its

motion (Fig. 1)? In answering this question the mathema-

tician begins exactly as does the policeman in setting a trap for

speeding. He takes a point Q near to P and determines the

distance PQ and the time it takes to pass over that distance.

Suppose, for example, that the distance PQ is ^ mi. and the

time is 1 min. Then, by 2, the average speed with which

the distance is traversed is

mi. = 30 mi. per hour.hr-

This is merely the average speed, however, and can no more

be taken for the true speed at the point P than could the 20 mi.

an hour which we obtained by considering the entire distance

A3. It is true that the 30 mi. an hour obtained from the

interval PQ is likely to be nearer the true speed at B than

was the 20 ini. an hour obtained from AS, because the interval

PQ is shorter.

The last statement suggests a method for obtaining a still

better measure of the speed at P ; namely, by taking the interval

PQ still smaller. Suppose, for example, that PQ is taken as

fa mi. and that the time is 6J sec. A calculation shows that the

average speed at which this distance was traversed was 36 mi.

an hour. This is a better value for the speed at P.

Now, having seen that we get a better value for the speed at

P each time that we decrease the size of the interval PQ, we

can find no end to the process except by means of the idea of a

limit denned in 1. We say, in fact, that the speed of a moving

body at any point of its path u the limit approached ly the average

6 RATES

speed computed for a small distance beginning at that point, the

limit to be determined by taking this distance smaller and smaller.

This definition may seem to the student a little intricate, and

we shall proceed to explain it further.

In the case of the automobile, which we have been using for

an illustration, there are practical difficulties in taking a very

small distance, because neither the measurement of the distance

nor that of the time can be exact. This does not alter ^the fact, however, that theoretically to determine the speed

of the car we ought to find the time it takes to go an

extremely minute distance, and the more minute the dis-

tance the better the result. For example, if it were possi-

ble to discover that an automobile ran ^ in. in-^^-g- sec.,

we should be pretty safe in saying that it was moving at pa speed of 30 mi. an hour. _p

Such fineness of measurement is, of course, impossible ;

but if an algebraic formula connecting the distance and

the time is known, the calculation can be made as fine as

this and finer. We will therefore take a familiar case in

which such a formula is known ; namely, that of a falling body.Let us take the formula from physics that if s is the distance

through which a body falls from rest, and t is the time it takes

to fall the distance st then

s = 16*3, (1)

and let us ask what is the speed of the body at the instant

when t SB 2. In Fig. 2 let be the point from which the bodyfalls, % its position when t 2, and 7 its position a short time

later. The average speed with which the body falls through the

distance PPZ is, by 2, that distance divided by the time it

takes to traverse it. We shall proceed to make several succes-

sive calculations of this average speed, assuming 7^ and the

corresponding time smaller and smaller.

In so doing it will be convenient to introduce a notation as

follows : Let ^ represent the time at which the body reaches 2?,

and ta the time at which it reaches J%. Also let ^ equal the

distance OP^ and a the distance OPa . Then s^-s^P^ and

SPEED 7

tf2 ^ is the time it takes to traverse the distance P^. Then the

average speed at which the body traverses Ufa is

So Si

(2)

Now, by the statement of our particular problem,

Therefore, from (1), s1= 16 (2)2= 64.

We shall assume a value of taa little larger than 2, compute

safrom (1), and the average speed from (2). That having been

done, we shall take tza little nearer to 2 than it was at first, and

again compute the average speed This we shall do repeatedly,

each time taking tanearer to 2.

Our results can best be exhibited in the form of a table, as

follows :

ta -L

2.1 70.56 .1 656 656201 646416 01 .6416 64.16

2 001 64.064016 .001 064016 64 016

2.0001 64 00640016 .0001 .00640016 64 0016

It is fairly evident from the above arithmetical work that as

the time i?

3 ^ and the corresponding distance ss

sxbecome

smaller, the more nearly is the average speed equal to 64.

Therefore we are led to infer, in accordance with 1, that the

speed at which the body passes the point J$ is 64 ft. per second.

In the same manner the speed of the body may be computedat any point of its path by a purely arithmetical calculation. In

the next section we shall go farther with the same problem and

employ algebra.

EXERCISES

1. Estimate the speed of a falling body at the end of the third

second, given that s = 16 tz, exhibiting the work in a table.

2. Estimate the speed of the body in Ex. 1 at the end of the

fourth, second, exhibiting the work in a table.

8 RATES

3. The distance of a falling body from a fixed point, at, any th

is given by the equation s = 100 + 16 t*. Estimate the speed of t

body at the end of the fourth second, exhibiting the work in a tab

4. A body is falling so that the distance traversed in the tiuu

is given by the equation s = 16 12 + 10 1 Estimate tho speed of 1,1

body when t = 2 sec, exhibiting the work in a table.

6. A body is thrown upward with such a speed that at. any tin

its distance from the surface of the earth is given by tho equsitus = 100 1 16 t

a. Estimate its speed at the end o a sexumd, exhibi

ing the work in a table.

6. The distance of a falling body from a fixed point a,t any tiin

is given by the equation s = 50 -f- 20 1 + 16 1*. Estimate its speeat the end of the first second, exhibiting tho work in a table.

4. Algebraic method. In this section we shall show how it i

possible to derive an algebraic formula for the speed, still con

fining ourselves to the special example of the falling body whos<

equation of motion is.,

, 2 ^s = i.o t .

(1

Instead of taking a definite numerical value for ft, wo dial

keep the algebraic symbol tr Then

Also, instead of adding successive small qimntit,ie,s to t to

get 2, we shall represent the amount added by the al

symbol^. That is, , .

,

.

and, from (1), sz== 16 (

Hence *2- ^=16^+ A)

2- 10 tf. 82

This is a general expression for the distance /tV* in Ffj.Now t

z-t

:=

Ti, and therefore the average speed withbody traverses %PS is represented by tho expression

It is obvious that if Ji is taken smaller and smidH Iho nver-age speed approaches 32 ^ as a limit, In

i'aufc, tho quantity 83^

ACCELEKATION 9

satisfies exactly the definition of limit given in 1. For if e

is any number, no matter how small, we have simply to take

10 Ti < e in order that the average speed should differ from 32 ^

by less than e;and after that, for still smaller values of h, this

d iff01 once remains less than e.

We have, then, the result that if the space traversed by a

1 ailing body is given by the formula

the speed of the body at any time is given by the formula

It may be well to emphasize that this is not the result which

would be obtained by dividing s by t.

EXERCISE

Find the speed in each of the problems in 3 by the method

explained in this section.

5. Acceleration. Let us consider the case of a body which is

supposed to move so that if s is the distance in feet and t is the

time in seconds, s = ts (V)

Then, by the method of 4, we find that if v is the speed in

feet per second, = 3 i2

We see that when t = 1, v = 3 ; when t = 2, v 12 ; when t = 3,

v = 27;and so on. That is, the body is gaming speed with each

second. We wish to find how fast it is gaining speed. To find

this out, let us take a specific time

(,-4.

The speed at this time we call vlt

so that, by (2),

i== 3 (4)

2= 48 ft. per second.

Take *2=5;

then v = 3 (5)2 ~ 75 ft. per second.

10 RATES

Therefore the body has gained 75 48 = 27 units of speed in

1 sec. This number, then, represents the average rate at which

the body is gaining speed during the particular second con-

sidered. It does not give exactly the rate at which the speed

is increasing at the beginning of the second, because the rate

is constantly changing.To find how fast the body is gaining speed when ^ = 4, we

must proceed exactly as we did in finding the speed itself.

That is, we must compute the gain of speed in a very small

interval of time and compare that with the time.

Let us take t.= 4.1.a

Then2= 50.43

and ^^=2.43.Then the body has gained 2.43 units of speed hi .1 sec., which

2 43is at the rate of -~- 24.3 units per second.

Again, take2=4.01.

Then v2= 48.2403

and ^^=.2403,A gain of ,2403 units of speed in .01 sec. is at the rate of

'

. =24.03 units per second. We exhibit these results, and

one other obtained in the same way, in a table:

4,1 60.43 .1 2.48 24.3

4.01 48.2403 .01 .2403 24,08

4.001 48.024003 001 .024003 24.003

The rate at which a body is gaining speed is called its

acceleration. Our discussion suggests that in. the example before

us the acceleration is 24 units of speed per second. But the

unit of speed is expressed in feet per second, and so we saythat the acceleration is 24 ft. per second per second.

KATE OF OHANG-E 11

By the method used in determining speed, we may get a

general formula to determine the acceleration from equation (2).

We take ,,

r

Then va=

and vz~Vi=

The average rate at which the speed is gamed is then

h,

and the limit of this, as h becomes smaller and smaller, is

obviously 6tt

.

This is, of course, a result which is valid only for the special

example that we are considering. A general statement of the

meaning of acceleration is as follows:

. , ,. ,. ., , change in speedAcceleration = limit of . 5

change in time

EXERCISES

1. If s = 4 1*, find the speed and the acceleration when t = #r2. If s = rf

8+ 1*, find the speed and the acceleration when t = 2.

3. If s = 3 tz + 2 1 + 5, how far has the body moved at the end

of the fifth second ? With what speed does it reach that point, and

how fast is the speed increasing?

4. If s = 4 i8 + 2 1

3+ 1 + 4, find the distance traveled and the

speed when t = 2.

5. If s = $ t6 + 1 -t- 10, find the speed and the acceleration when

t=z>2 and when t = 3. Compare the average speed and the average

acceleration during this second with the speed and the acceleration

at the beginning and the end of the second.

6. If s = at + &, show that the speed is constant.

7. If at2 + to + G, show that' the acceleration is constant.

8. If s = at* + bt*+ ct +/, find the formulas for the speed and

the acceleration.

6. Rate of change. Let us consider another example which

may be solved by processes similar to those used for determining

speed and acceleration.

L2 KATES

m the

A stone is thrown into still water, forming ripples which

,ravel from the center of disturbance in the form of circles

Tig. 3). Let r be the radius of a circle and A its area. ThenJ "J /-I NA=irr. (1)

We wish to compare changes in the area with changesadius. If we take r

t= 3, then A^ 9 TT; and if we take

,hen ^ta= 167T. That is, a change

)f 1 unit in ?*, when r 3, causes

i change of 7 TT units in A. We are

^empted to say that A is increas-

ng TTT times as fast as r. But

jefore making such a statement it

s well to see whether this law holds

'or all changes made in r, starting

'rom r1= 8, and especially for small

jhanges in r.

We will again exhibit the calcu- ^IQ 3ation in the form of a table. Here

^=3, A^Qnr, and rz

is a variously assumed value of r not

nuch different from 3.

1

01

.001

9.61 TT

9.0601 TT

9,006001 f

.1

.01

001

A,,- A!

.01 TT

.0001 7T

000001 IT

ra- r

x

ITT

6.01 TT

6 001 TT

The number in the last column changes with the number

a r^ Therefore, if we wish to measure the rate at which A is

3hanging as compared with r at the instant when r = 3, we mustbake the limit of the numbers in the last column. That limit is

abviously 6 TT.

We say that at the instant when r 8, the area of the circle is

changing 6 TT times as fast as the radius. Hence, if the radius

Is changing at the rate of 2 ft. per second, for example, the area

is changing at the rate of 12 TT sq. ft. per second. Another wayof expressing the saine idea is to say that when, r ts 3, the rate

RATE OF CHANGE 13

of change of A with respect to r is 6 TT. Whichever form of expres-sion is used, we mean that the change in the area divided bythe change in the radius approaches a limit 6 TT.

The number 6 TT was, of course, dependent upon the value

r = 3, with which we started. Another value of ^ assumed at

the start would produce another result. For example, we maycompute that when r

x= 4, the rate of change of A with respect

to r is 8-rr; and when r^= 5, the rate is 10 ?r. Better still, we

may derive a general formula which will give us the required

rate for any value ofr^.

To do this take

Then A2= TT (r*+ 2 rji + A2

)

and Az- A

1= TT (2 r

xh + 7i

3

) ;

A ASO that 2 l = 2 TH* + 7i7T.

T V'z 'l

The limit of this quantity, as Ti is taken smaller and smaller, is

Hence we see that from formula (1) we may derive the fact

that the rate of change of A with respect to r is 2 TIT.

EXERCISES

1. In the example of the text, if the circumference of the circle

winch bounds the disturbed area is 10 ft and the circumference is

increasing at the rate of 3 ft. per second, how fast is the area

increasing ?

2. In the example of the text find a general expression for the

rate of change of the area with respect to the circumference.

3. A soap bubble is expanding, always remaining spherical. If

the radius of the bubble is increasing at the rate of 2 in. per second,

how fast is the volume increasing ?

4. In Ex 3 find the general expression for the rate of change

of the volume with respect to the radius.

5. If a soap bubble is expanding as in Ex. 3, how fast is the

area of its surface increasing9

14 BATES

6. In Ex. 5 find the general expression for the rate of change

of the surface with respect to the radius

7. A cube of metal is expanding under the influence of heat.

Assuming that the metal retains the form of a cube, find the rate of

change at which the volume is increasing with respect to an edge.

8. The altitude of a right circular cylinder is always equal to

the diameter of the base. If the cylinder is assumed to expand,

always retaining its form and proportions, what is the rate of change

of the volume with respect to the radius of the base ?

9. Find the rate of change of the area of a sector of a circle of

radms 6 ft with respect to the angle at the center of the circle.

10. Find the rate of change of the area of a sector of a circle

with respect to the radius of the circle if the angle at the center

7Tof the circle is always j. What is the value of the rate when the

radius is 8 in. ?

CHAPTER II

DIFFERENTIATION

7. The derivative. The examples we have been consideringin the foregoing sections of the book are alike in the methods

used to solve them. We shall proceed now to examine this

method so as to bring out its general character.

In the first place, we notice that we have to do with two

quantities so related that the value of one depends upon the

value of the other. Thus the distance traveled by a moving

body depends upon the time, and the area of a circle depends

upon the radius. In such a case one quantity is said to be

a function of the other. That is, a quantity y is said to be a

function of another quantity, x, if the value of y is determined ly

the value of x.

The fact that y is a function of x is expressed by the equation

y=/<v>,

and the particular value of the function when x has a definite

value a is then expressed as /(a). Thus, if

f(x) = x*- 3 aM-42j + l,

/(2)=2-3(2)3

+4(2)+ l = 5,

/(0)=0-3(0) + 4(0) + 1= 1.

It is in general true that a change in x causes a change in

the function y, and that if the change in x is sufficiently small,

the change in y is small also. Some exceptions to this may be

noticed later, but this is the general rule. A change in a; is

called an increment of x and is denoted by the symbol Ao? (read"delta x "). Similarly, a change in y is called an increment of

y and is denoted by Ay. For example, consider

15

16 DIFFERENTIATION

When = 2, # = 12. When a? =2.1, y = 12.71. The changein x is .1, and the change m y is .71, and we write

Aa= .l, Ay= .71.

So, in general, if a^ is one value of a;, and xza second value

of #, thenAx = xs,

ixl1

or #2=0^4- Are; (1)

and if yxand y2

are the corresponding values of y, then

ty=yz-Vv <* y2=ya4Ay. (2)

The word increment really means "increase," but as we are

dealing with algebraic quantities, the increment may be nega-tive when it means a decrease. For example, if a man invests

$1000 and at the end of a year has $1200, the increment of his

wealth is $200. If he has $800 at the end of the year, the

increment is $200. So, if a thermometer registers 65 in the

morning and 57 at night, the increment is ,8. The incre-

ment is always the second value of the quantity considered minus

the first value.

Now, having determined increments of x and of y, the next

step is to compare them by dividing the increment of y bythe increment of x. This is what we did in each of the three

problems we have worked in 3-6. In finding speed we began

by dividing an increment of distance by an increment of time,

in finding acceleration we began by dividing an increment of

speed by an increment of time, and in discussing the ripples in

the water we began by dividing an increment of area by an

increment of radius. .

The quotient thus obtained is - That is,Ax

Ay _ increment of y _ change in yAJC increment of x change m x

An examination of the tables of numerical values in 3, 5, 6

shows that the quotient^depends upon the magnitude of Asc,

and that in each problem it was necessary to determine its limit

DEEIVATIVE 17

as A# approached zero. This limit is called the derivative of y

with respect to x, and is denoted by the symbol-~- We have then

fy v -j. Ay v , change in v~ = limit oi ~ = limit of . .

a

ax Ax change in x

At present the student is to take the symbol-^ not as adx

fraction, but as one undivided symbol to represent the deriva-

tive. Later we shall consider what meaning may be given to

dx and dy separately. At this stage the form -^ suggests simplyA (%>*Xs

the fraction ^, which has approached a definite limiting value,

The process of finding the derivative is called differentiation,

and we are said to differentiate y with respect to x. From the

definition and from the examples with which we began the book,

the process is seen to involve the following four steps :

1. The assumption at pleasure of Ax.

2. The determination of the corresponding Ay.Aw

8. The division of Ay by Aa: to form -*-Ax

4. The determination of the limit approached by the quotient

in step 3 as the increment assumed in step 1 approaches zero.

Let us apply this method to finding-J*-

when y - - Let xt

be a definite value of x, and y,- the corresponding value of y.x

i

1. Take Ax = 7i.

Then, by (1), x^x^h.

2. Then ^l^-J.;

11 hwhence, by (2), Ay-

3. By division, T^ = . . 7J Ax asf+AiS

18 DIEFEBENTIATION

4. By inspection it is evident that the limit, as h approaches

zero, is 5 which is the value of the derivative when x-=x*.xl

But x^ may be any value of x ; so we may drop the subscript 1 and

write as a general formula

dx 3?

EXERCISES

Pind from the definition the derivatives of the following ex-

pressions :

6. y-rf + i.

22. y = *+2a*+i. e

-y=2+*3. y = a:

4 ~a8. 7. y = "+ \x*+ x - 5.

i =i 8 = 3a;2 + 1.

y ~~xs y ~~

x

8. Differentiation of a polynomial. We shall now obtain for-

mulas by means of which the derivative of a polynomial may be

written down quickly. In the first place we have the theorem :

The derivative of a polynomial is the sum of the derivatives of

its separate terms.

This follows from the definition of a derivative if we reflect

that the change in a polynomial is the sum of the changes in its

terms. A more formal proof will be given later.

We have then to consider the terms of a polynomial, which

have in general the form #af. Since we wish to have general

formulas, we shall omit the subscript 1 in denoting the first

values of x and y. We have then the theorem:

If y =s ax*) where n is a positive integer and a is a constant, then

(1)dx ^ '

To prove this, apply the method of 7 :

1. Take Aa?=A;whence jc

a= x + h.

POLYNOMIAL 19

2. Then y3=ax2 = a(z + A)";

whence Ay == a (x + A)" are"

.4

3. By division,^ = a(nx^-l+

n ^n ~ 1^ tf-'h + . . . + A"" 1

)-ZA3/ j >

4. By inspection, the limit approached by > as Ji approaches

zero, is seen to be anxn~ l

.

Therefore ~ = anxn~\ as was to be proved.dx

The polynomial may also have a term of the form ax. This

is only a special case of (1) with w = l, but for clearness wesay explicitly:

If y = ax, where a is a constant, then

- (2)dx

Finally, a polynomial may have a constant term. c. For this

we have the theorem:

If y = G, where c is a constant, then

f^O. (3)ax

The proof of this is that as c is constant, A<? is always zero,

no matter what the value of A is. Hence

and therefore = 0.ax

As an example of the use of the theorems, consider

yWe write at once

ax

20 DIFFERENTIATION

EXERCISES

Find the derivative of each of the following polynomials :

1. 3.2 + a- 3. 6. x 1 + 7 xs+ 21 8 - 14 as

2. o;8 +2a; + l. 7. x*- x* + 4a - 1

3. o:4 + 4a58 +6a;2 +4a; + l. 8. 3

4. x6 + |a:4 +2a;2 +3. 9. aa

5. a;6 -4a:4 + a:

2 -4a; 10. a + bxz + ca;4 + ex6

.

i

9. Sign of the derivative. If ~ is positive, an increase in thedx

, dy .

value of x causes an increase in the value of y. If~ is negative,dx

an increase in the value of x causes a decrease in the value of y.

To prove this theorem, let us consider that-jj-

is positive.

Then, since ~- is the limit of -> it follows that is positivedx A A

for sufficiently small values of Aa;; that is, if A# is assumed

positive, ky is also positive, and therefore an increase of x

causes an increase of y. Similarly, if-j-

is negative, it follows

Awthat is negative for sufficiently small values of Aa: ; that is,

if Arc is positive, A?/ must be negative, so that an increase of xcauses a decrease of y.

In applying this theorem it is necessary to determine the

sign of a derivative. In case the derivative is a polynomial, this

may be conveniently done by breaking it up into factors and

considering the sign of each factor. It is obvious that a factor

of the form x a is positive when x is greater than a, and

negative when x is less than a.

Suppose, then, we wish to determine the sign of

There are three factors to consider, and three numbers are im-

portant ; namely, those which make one of the factors equal to

zero. These numbers arranged in order of size are 3, 1, and 6.

We have the four cases :

1. x< 3. All factors are negative and the product is

Dative.

VELOCITY AND ACCELERATION 21

2. 3 < x < 1, The first factor is positive and the others

are negative. Therefore the product is positive.

3. 1 < x < 6. The first two factors are positive and the last

is negative. Therefore the product is negative.

4. x > 6. All factors are positive and the product is positive.

As an example of the use of the theorem, suppose we have

# = Xs- 3;y*-Q x + 27,

and ask for what values of x an increase in y will cause an

increase in y. We form the derivative and factor it. Thus,

^==3aa

-6a3-9==3(a: + r)(a;-3).//*

Proceeding as above, we have the three cases:

1. x< 1. -^ is positive, and an increase in x thereforedx

increases y. _

2. 1 < x < 3. -^ is negative, and therefore an increase in x, dxdecreases y. _

3. a; > 3. -^ is positive, and therefore an increase in x in-

creases y.

These results may be checked by substituting values of x in

the derivative.EXERCISES

Pind for what values of x each of the following expressions will

increase if x is increased, and for what values of x they will decrease

if x is increased :

1. a!> -4aj + 6. 6.

2. 3aa +10a; + 7. 7. xa - x*- 5x + 5

3. l + Ssc-a:2. 8. 1 + 6x +12a:2 + 8oj8.

4. 7_3a._3a;2 9. 6 + 60; + 6aa - 2 3 - 3

5. 2 je8 +3a;a - 12 + 17. 10. 12 - 12a - 6a;

2 + 4a;8 +

10. Velocity and acceleration (continued). The method bywhich the speed of a body was determined in 4 was in reality

a method of differentiation, and the speed was the derivative of

the distance with respect to the time. In that discussion, how-

ever, we SQ arranged each problem that the result was positive

22 DIFFERENTIATION

and gave a numerical measure (feet per second, miles per hour,

etc.) for the rate at which the body was moving. Since we maynow expect, on occasion, negative signs, we will replace the word

speed by the word velocity, which we denote by the letter v.

In accordance with the previous work, we have

da -, N

* a)

The distinction between speed and velocity, as we use the

words, is simply one of algebraic sign. The speed is the numer-

ical measure of the velocity and is always positive, but the

velocity may be either positive or negative.

From 9 the velocity is positive when the body so moves that

s increases with the time. This happens when the body moves in

the direction in which s is measured. On the other hand, the

velocity is negative when the body so moves that s decreases

with the time. This happens when the body moves in the direc-

tion opposite to that in which s is measured.

For example, suppose a body moves from A to B (Fig. 1), a

distance of 100 mi., and let P be the position of the body at a

time t, and let us assume that we know that AP = 4 1. If wemeasure s from A, we have

dswhence v = -= 4.

dt

On the other hand, if we measure s from J5, we have

whence v - = 4.dt

"We will now define acceleration by the formula

dv

*=win full accord with 5 ; or, since v is found by differentiating s,

we may write

VELOCITY AND ACCELERATION 23

where the symbol on the right indicates that * is to be differ-

entiated twice in succession. The result is called a second

derivative,

A positive acceleration means that the velocity is increasing,

but it must be remembered that the word increase is used in

the algebraic sense. Thus, if a number changes from 8 to

5, it algebraically increases, although numerically it decreases.

Hence, if a negative velocity is increased, the speed is less. Simi-

larly, if the acceleration is negative, the velocity is decreasing,

but if the velocity is negative, that means an increasing speed.

There are four cases of combinations of signs which mayoccur :

1. v positive, a positive. The body is moving in the direction

in which s is measured and with increasing speed.

2. v positive, a negative. The body is moving in the direction

in which s is measured and with decreasing speed.

8. v negative, a positive. The body is moving in the direction

opposite to that in which s is measured and with decreasing

speed.

4. v negative, a negative. The body is moving in the direc-

tion opposite to that in which 8 is measured and with increasing

As an example, suppose a body thrown vertically into the air

with a velocity of 96 ft. per second. From physics, if s is meas-

ured up from the earth, we have

From this equation we compute

v = 96 - 32 t,

When t< 3, v is positive and a is negative. The body is going

up with decreasing speed. When t > 3, v is negative and a is

negative. The body is coming down with increasing speed.

On the other hand, suppose a body is thrown down from a

height with a velocity of 96 ft. per second. Then, if 8 is measured

24 DIFFERENTIATION

down from the point from which the body is thrown, we have,

from physics,8 = Wt+lQt*,

from which we computev = 96 + 32 1,

= 32.

Here v is always positive and a is always positive. There-

fore the hody is always going down (until it strikes) with an

increasing speed.EXERCISES

In the following examples find the expression for the velocity

and determine when the body is moving in the direction in which

s is measured and when in the opposite direction .

1. s = t* 3t + 6. 3. s = t* Qt* + 24* + 3.

2. s = 10* - tf 4. 's = 8 + 12* - 6i2 + <8.

5. s = 254 -

In the following examples find the expressions for the velocityand the acceleration, and determine the periods of time during whichthe velocity is increasing and those during which it is decreasing .

6. s = 3z5-4i + 4 8. s = %t* 2 2

7. s = 1 + 5t - #. 9. s = t* - 5t* + St + 1.

10. s = 1 + 4 + 2t* t8

.

11. Rate of change (continued). In 6 .we have solved a

problem in which we are finally led to find the rate of increase

of the area of a circle with respect to its radius. This problemis typical of a good many others.

Let x be an independent variable and y a function of x.

A change Aa; made in x causes a change Ay in y. The fraction

compares the change in y with the change in x. For exam-

ple, if Aa; = .001, and Ay = .009061, then we may say that the

change in y is at the average rate of' = 9.061 per unit

change in x. This does not mean that a unit change in a? would

actually make a change of 9.061 units in y, any more than the

RATE OF CHANGE 25

statement that an automobile is moving at the rate of 40 mi. anhour means that it actually goes 40 mi. in an hour's time.

The fraction then gives a measure for the average rate atwhich y is changing compared with the change in x. But this

measure depends upon the value of A#, as has been shown inthe numerical calculations of 6. To obtain a measure of theinstantaneous rate of change of y with respect to x which shall

not depend upon the magnitude of A#, we must take the limit

of ~ t as we did in G.AxWe have, therefore, the following definition :

Tlie derivative-j-

measures the rate ofchange ofy with respect to a\ax

Another way of putting the same thing is to say that if -^has the value m, then y is changing m times as fast as x.

Still another way of expressing the same idea is to say that

the rate of change of y with respect to x is defined as meaningthe limit of the ratio of a small changein y to a small change in x.

We will illustrate the above general

discussion, and at the same time show

how it may be practically applied, by the

following example, which we will first

solve arithmetically and then by calculus.

Suppose we have a vessel in the shape of

a cone (Fig 4) of radius 3 in. and altitude

9 in. into which water is being poured at

the rate of 100 cu. in. per second. Re-

quired the rate at which the depth of the

water is increasing when the depth is 6 in.

From similar triangles in the figure, if h is the depth of the

water and r the radius of its surface, r -. If Fis the volumeof water, w , . i , ,^

FlG. 4

We are asked to find the rate at which the depth is increasingwhen h is C in. Let us call that depth so that 7^= 6. Then

26 DIFFERENTIATION

y = 8 TT. Now we will increase \ by successive small amounts

and see how great an increase in V^ is necessary to cause that

change in \ ;that is, how much water must be poured in to raise

the depth by that amount. The calculation may be tabulated as

follows .

A h AF ATA /i

.1 .407 IT 4 07 ir

01 04007 TT 4.007 v.001 0040007 TT 4.0007 TT

The limit of the numbers in the last column is evidently 4 TT.

Therefore the volume is increasing 4 IT times as fast as the

depth. But, by hypothesis, the volume is increasing at the rate

of 100 cu. in. per second, so that the depth is increasing at the

rate of - = 7.96 in. per second.47T

We have solved the problem by arithmetic to exhibit againthe meaning of the derivative. The solution by calculus is muchquicker. We begin by finding

dV 1 , 3

aTe**'This is the general expression for the rate of change of V

with respect to A, or, in other words, it tells us that V is instan-

taneously increasing \ rf times as fast as li for any given A.

Therefore, when A = 6, V is increasing 4 TT times as fast as A,

and as V is increasing at the rate of 100 cu. m. per second, A is

increasing at the rate of - = 7.96 in. per second.4 7T

EXERCISES

1. An icicle, which is melting, is always in the form of a rightcircular cone of which the vertical angle is 60 Find the rate of

change of the volume of the icicle with respect to its length.2. A series of right sections is made in a right circular cone of

which the vertical angle is 90 How fast will the areas of the sec-

tions be increasing if the cutting plane recedes from the vertex atthe rate of 3 ft. per second ?

GRAPHS 27

3. A solution is being poured into a conical filter at the rate of

5 cc per second and is running out at the rate of 1 ce. per secondThe radius of the top of the filter is 10 cm. and the depth of the

filter is 30 cm Find the rate at which the level of the solution is

rising in the filter when it is one fourth of the way* to the top.

4. A peg in the form of a right circular cone of which the ver-

tical angle is 60 is being driven into the sand at the rate of 1 m.

per second, the axis of the cone being perpendicular to the surface

of the sand, which is a plane. How fast is the lateral surface of the

peg disappearing in the sand when the vertex of the peg is 5 in.

below the surface of the sand?

5. A trough is in the form of a right prism with its ends equi-lateral triangles placed vertically The length of the trough is 10 ft

It contains water which leaks out at the rate of cu ft per minute.

Find the rate, in inches per minute, at which the level of the water

is sinking in the trough when the depth is 2 ft.

6. A trough is 10 ft. long, and its cross section, which is vertical,

is a regular trapezoid with its top side 4ft. in length, its bottom

side 2 ft,and its altitude 5 ft. It contains water to the depth of

3 ft,and water is running in so that the depth is increasing at the

rate of 2 ft. per second. How fast is the water running in ?

7. A balloon is in the form of a right circular cone with a hemi-

spherical top. The radius of the largest cross section is equal to

the altitude of the cone. The shape and proportions of the balloon

are assumed to be unaltered as the balloon is inflated. Find the

rate of increase of the volume with respect to the total height of

the balloon.

8. A spherical shell of ice surrounds a spherical iron ball concen-

tric with it. The radius of the iron ball is 6 in. As the ice melts,

how fast is the mass of the ice decreasing with respect to its thickness ?

12. Graphs. The relation between a variable x and a function

y may be pictured to the eye by a graph. It is expected that

students will have acquired some knowledge of the graph in

the study of algebra, and the following brief discussion is givenfor a review.

Take two lines OX and 07 (Fig. 5), intersecting at right

angles at 0, which is called the origin of codrdinates. The line

called the axis of as, and the line 0rthe axis of y ; together

28 DIFFERENTIATION

they are called the coordinate axes, or axes of reference. On OXwe lay off a distance OM equal to any given value of x, measur-

ing to the right if x is positive and to the left if x is negative.

From M we erect a perpendicular MP, equal in length to the

value of y, measured up if y is positive and down if y is negative.

The point P thus determmed is said to have the coordinates

x and y and is denoted by (x, y). It follows that the numerical

value of x measures the distance Yof the point P from OY, and the

numerical value of y measures the

distance of P from OX. The coor-

dinate x is called the abscissa, and

the coordinate y the ordinate. It is p

evident that any pair of coordinates

(x, y) fix a single point P, and that

any point P has a single pair of M -^

coordinates. The point P is said to _,g

be plotted when its position is fixed

111 this way, and the plotting is conveniently carried out on

paper ruled for that purpose into squares.If y is a function of x, values of x may be assumed at pleasure

and the corresponding values of y computed. Then each pair of

values (x, y) may be plotted and a series of points found. Thelocus of these points is a curve called the graph of the function.

It may happen that the locus consists of distinct portions not

connected in the graph. In this case it is still customary to saythat these portions together form a single curve.

For example, let - ... _r y~bz ar. (1)

We assume values of x and compute values of y. The results

are exhibited in the following table :

-i o l 2 8 4 6'

o

2/-60 4 6 6 4 0-0

These points are plotted and connected by -a Smooth curve*

giving the result shown in F-ig. 6. This -curve should have the

GEAPHS 29

property that the coordinates of any point on it satisfy equa-tion (1) and that any point whose coordinates satisfy (1) lies

on the curve. It is called the graph both of the function y and

of the equation (1), and equation (1) is called the equation of

the curve.

Of course we are absolutely sure of only those points whose

coordinates we have actually computed. If greater accuracy is

desired, more points must be found yby assuming fractional values of x.

For instance, there is doubt as to the 7.

shape of the curve between the points6 / \

(2, 6) and (3, C). We take, therefore, 4' >

#=2^- and find /=

6|. This gives 3(

us another point to aid us in draw- 2

ing the graph. Later, by use of the 1

calculus, we can show that this last -1 0^1 2 8 4 5

point is really the highest point of"

-2

the curve. -s

The curve (Fig. 6) gives us a~4

graphical representation of the way"^

in which y varies with x. We see, for

example, that when x varies from 1

to 2, y is increasing; that when x

varies from 3 to 6, y is decreasing; and that at some point

between (2, 6) and (3, 6), not yet exactly determined, y has its

largest value.

It is also evident that the steepness of the curve indicates in

some way the rate at which y is increasing with respect to x.

For example, when # = 1, an increase of 1 unit in x causes an

increase of 6 units in y ; while when a =1, an increase of 1 unit

in x causes an increase of only 2 units m y. The curve is

therefore steeper when x = 1 than it is when x = 1.

Now we have seen that the derivative - measures the rateax

of change of y with respect to x. Hence we expect the derivative

to be connected in some way with the steepness of the curve.

We shaft tihmwfnro /iianr.. *Hr. ^QTL m gs \%. and" 15.

USc Lib B'lore Q515N22 ! 13838in IIIIIIIIIIIIIIIIIMIII ~

30 DIFFERENTIATION

1

EXERCISES

Plot the graphs of the following equations :

1. y = 2x -|-3. 4. ?/= a;

2 -5a; + 6. 7. y = zc8

.

2. y = 2x + 4. 5. y = a;2 + 4 a; + 8 8. y = x* 4aj

a

3. y 5. 6. y 9 3 a; a;2

. 9.j/= a:

8 1.

10. What is the effect on the graph of y mx + 3 if different

values are assigned to m ? How are the graphs related ? Whatdoes this indicate as to the meaning of m ?

11. What is the effect on the graph of y = 2x + J if different

values are assigned to b ? What is the meaning of & ?

12. Show by similar triangles that y = mx is always a straightline passing through 0.

13. By the use of Exs. 11 and 12 show that y = mx + b is alwaysa straight line

13. Real roots of an equation. It is evident that the real roots

of the equation f(x) determine points on the axis of x at

which the curve y=f(x) crosses or touches that axis. More-

over, if x1and

2 ( t<*2)are two values of x such that f(x^)

and/( 2) are of opposite algebraic sign, the graph is on one side

of the axis when x = xi ,

and on the other side when ic = a;2

.

Therefore it must have crossed the axis an odd number of times

between the points x = zland x = x

z. Of course it may have

touched the axis at any number of intermediate points. Now, if

f(x) has a factor of the form (x a)1, the curve y =/(#) crosses

the axis of x at the point x = a when k is odd, and touches the

axis of x when k is even. In each case the equation /(#) = is

said to have Jc equal roots, x a. Since, then, a point of crossingcorresponds to an odd number of equal roots of an equation, and a

point of touching corresponds to an even number of equal roots,it follows that the equation f(x) = has an odd number of real

roots between x^ and xz

if /(#,) and / (:c2) have opposite signs.The above gives a ready means of locating the real roots of

an equation in the form /(z) = 0, for we have only to find twovalues of x, as x

land a?

2 , for which f(x) has different signs. Wethen know that the equation has an odd number of real roots

STEAIGHT LINE 31

between these values, and the nearer together xland #

2 ,the

more nearly do we know the values of the intermediate roots.

In locating the roots in this manner it is not necessary to con-

struct the corresponding graph, though it may be helpful.

Ex. Find a real root of the equation Xs + 2 x 17 = 0, accurate to two

decimal places.

Denoting x3 + 2 x 17 by f(x) and assigning successive integial values

to x, we find/(2) = 5 and/(3) = 16 Hence there is a leal root of the

equation between 2 and 3.

We now assign values to x between 2 and 3, at intervals of one tenth,

as 2 1, 2 2, 2.3, etc., and we begin with the values nearei 2; since /(2) is

nearer zeio than is/(3). Proceeding in this way we find jf (2.3)= 233

and/(2 4) = 1.624, hence the root is between 2.3 and 2.4.

Now, assigning values to a? between 2 3 and 2.4 at intervals of one hun-

diedth, we find /(2.31) = - .054 and /(2.32) = 127, hence the root as

between 2 31 and 2.32.

To determine the last decimal place accurately, we let x = 2.315 and

find /(2.315) = .037. Hence the root is between 2 31 and 2 315 and is

2.31, accurate to two decimal places

If /(2 315) had been negative, we should have known the root to be

between 2 315 and 2.32 and to be 2 32, accurate to two decimal places.

EXERCISES

Find the real roots, accurate to two decimal places, of the follow-

ing equations :

1. a8 +2a;-6 = 0. 4. j*- 4a8+ 4 = 0.

2. a8 + o;+ll = 5. a:8 - 3ic

2+ 60: - 11 = 0.

3. *-lla: + 6 = 0. 6.

14. Slope of a straight line. Let LK (Figs. 7 and 8) be any

straight line not parallel to OX GS OY, and let J? (a^, y^ and

P2 (#2 , y3) be any two points on it. If we imagine a point to

move on the line from P1 to P^ the increment of x is #2x and

the increment of y is y^y^ We shall define the slope as the

ratio of the increment ofy to the increment ofx and denote it by m.

We have then, by definition,

82 DIFFERENTIATION

A geometric interpretation of the slope is easily given. For

if we draw through JFJa line parallel to OJf, and through J^ a

line parallel to OF, and call R the intersection of these lines,

then xz

xl= P

1E and y^y^RP^ Also, if $ is the angle

which the line makes with OX measured as in the figure, then

: tan 0. C )

It is clear from the figures as well as from formula (2) that

the value of m is independent of the two points chosen to define

it, provided only that these are on the given line. We may there-

fore always choose the two points so that y^y^ is positive.

EIG. 7 FIG. 8

Then if the line runs up to the right, as in Fig. 7, #2 a^ is

positive and the slope is positive. If the line runs down to the

right, as in Fig. 8, xz o^is negative and m is negative. There-

fore the algebraic sign of m determines the general direction in

which the line runs, while the magnitude of m determines the

steepness of the line.

Formula (1) may be used to obtain the equation of the line.

Let m be given a fixed value and the point I^(x^ y^) be held

fixed, but let Ji^ be allowed to wander over the line, taking on,

therefore, variable coordinates (a?, y). Equation (1) may then

be written y-y^m(x- ^). (3)

This is the equation of a line through a fixed point (a^, y^with a fixed slope m, since it is satisfied by the coordinates of

any point on the line and by those of no other point.

STRAIGHT LINE 33

In particular, P^ (x^ y^) may be taken as the point with coor-

dinates (0, i) in which the line cuts OY. Then equation (3)

becomesy = mx + l. (4)

Since any straight line not parallel to OX or to OY intersects

OY somewhere and has a definite slope, the equation of anysuch line may be written in the form (4).

It remains to examine lines parallel either to OX or to OF. If

the line is parallel to OX, we have no triangle as in Figs. 7 and 8,

but the numerator of the fraction m (1) is zero, and we there-

fore say such a line has the slope 0. Its equation is of the form

y = , (5)

since it consists of all points for which this equation is true.

If the line is parallel to OY, again we have no triangle as in

Figs. 7 and 8, but the denominator of the fraction in (1) is zero,

and m accordance with established usage we say that the slope of

the line is infinite, or that m=oo. This means that as the position

of the line approaches parallelism

with OY the value of the fraction

(1) increases without limit. The

equation of such a line is

x = a. (6)

Finally we notice that any equa-

tion of the form

^ +5y + (7=0 (7)

always represents a straight line. This follows from the fact that

the equation may be written either as (4), (5), or (6).

Tbe line (7) may be plotted by locating two points and

drawing a straight line through them. Its slope may be found

by writing the equation in the form (4) when possible. The

coefficient of x is then the slope.

If two lines are parallel they make equal angles with OX.

Therefore, if mland m

zare the slopes of the lines, we have,

from (2), mz=m,. (8)

If two lines are perpendicular and make angles <^ and ^>s

respectively with OX, it is evident from Fig. 9 that <

a= 90+ ^ ;

34 DIFFERENTIATION

whence tan $ == ~ cot $ = Hence, if m^ and maare th

slopes of the lines, we have

It is easy to show, conversely, that if equation (8) is satis

fied by two lines, they are parallel, and that if equation (9) i

satisfied, they are perpendicular. Therefore equations (8) am

(9) are the conditions for parallelism and perpendiculant

respectively.

Ex. 1. Find the equation of a straight line passing through the pom(1, 2) and parallel to the straight line determined by the two points (4, 2

and (2,-

3)

By (1) the slope of the line determined by the two points (4, 2) am_ g _ 2 5

(2, 3) is = - Therefore, by (3), the equation of the requiiei

line is n R / i\y- 2 = i (a;

-1),

which i educes to x 2y 1 = 0.

Ex. 2. Find the equation of a straight line through the point (2, 3

and perpendicular to the line 2 a: 3?/ + 7 =

The equation of the given straight line may be written in the forn

y = 3 x + , which is form (4). Therefore m = . Accordingly, by (0)the slope of the required line is By (3) the equation of the requimline is

y + 3 = -|(a;-2),

which reduces to 3 x + 2 y - 0.

Ex. 3. Find the equation of the straight line passing through the poinl

( 3, 3) and the point of intersection of the two lines 2 a? y 3 = anr

The coordinates of the point of intersection, of the two given lines musl

satisfy the equation of each line. Therefore the codrdinates of the poinlof intersection are found by solving the two equations simultaneously,The result is (1,

-1)

We now have the problem to pass a straight line through the points

(- 3, 3) and (1,-

1). By (1) the slope of the required line isJ*

+ 1 = - 1.

Therefore, by (3), the equation of the line is

which reduces to x + y = 0.

STRAIGHT LINE 35

EXERCISES

1. Find the equation of the straight line which passes through

(2,-

3) with the slope 3

2. Find the equation of the straight line which passes through( 3, 1) with the slope $

3. Find the equation of the straight line passing through the

points (1, 4) and (f , ).

4. Find the equation of the straight line passing through the

points (2, 3) and (- 3,-

3).

5. Find the equation of the straight line passing through the

point (2, 2) and making an angle of 60 with OX.

6. Find the equation of the straight line passing through the

point Q, |) and making an angle of 135 with OX.

^7. Find the equation of the straight hue passing through the

point ( 2, 3) and parallel to the line x + 2 y + 1 = 0.

8. Find the equation of the straight line passing through the

point ( 2, 3) and perpendicular to the line 3 a; + 4 ?/ 12 =0.9. Find the equation of the straight line passing through the

point (, )and parallel to the straight line determined by the two

points (f, |) and (J,-

)

10. Find the equation of the straight line passing through

(i~

i) an(^ perpendicular to the straight line determined by the

points (2, 1) and (- 3, 5).

11. If /? is the angle between two straight lines which make angles

#j and <2 (<j!>2 > <j) respectively with OX, prove from a diagram similar

to Fig 9 that /3=

<f>2< r If tan

t= m^ and tan

tj>z= w

a , prove bytrigonometry that

tan ft=

.f

l

1 + m2Wj

12. Find the angle between the lines # 2?/-fl=;0 and

13. Find the angle between the lines 2x 4^ + 5 = and

y 6= 0.

14. Find the angle between the lines y= Sas+4 and a?H-3y-f-7=0.15. The vertex of a right angle is at (2, 4) and one of its sides

passes through the point (- 2, 2). Find the equation of the other side.

16. Find the foot of the perpendicular from the origin to the line

36 DIFFERENTIATION

15. Slope of a curve. Let An (Fig. 10) be any curve serving

as the graphical representation of a function of a\ Let 1\ be any

point on the curve with coordinates y;l= OM^ t//,= yl/j/J. TU\o

l^xM^ and draw the perpendicular J/a /j, fixing the point /a

'

on the curve with the coor-

M in Q^OQ *T* (iT(/T Q/ " 1\/T T* * s OLiiixajUco^jj""""

'-'' 2* ,yjj"""

-*'*j*j" jr ^r>

Draw ^J? parallel to OX. /RThen

PR W.M"..= ArK. pV" -K

7jjJ5 = Jlf.,Pa M, 11= A?/, ^

\ a , r>t> /and

Ax* /

Draw the straight line / ^ ^j, prolonging it to form a JTIU . to

secant J? Then, by 14, -^ is the slope of the secant J>8, and

may be called the average slope of tho curve between the pointsJJ and PvTo obtain a number which may be used for tho actual slope

of the curve at the point / it is necessary to uso tho limit

process (with which the student should now bo familiar), bywhich we allow kx to become smaller and smaller and the pointPz to approach Pl along the curve. Tho result is tho derivativeof y with respect to #, and we have tho following result :

The slope of a curve at any point u yiven, ly the value of tfo

derivative -^ at that point.dx f

As this limit process takes place, tho point / approaching t!u

point Pyit appears from the ligure thai, the secant 7A' approaches

a limiting position 1{T. The line 1\T is called a tanymt to 1,lu>

curve, a tangent Iriny then by deflnitwn the Una approached an tt

limit by a secant throuc/h two point* of the curve a tht* two point*approach coincidence. It follows that tho slope* of tho tangent is

the limit of the slope of the secant, Therefore,

slope of a curve at any point in the mme an the dope of tfta

*angent at that point,

SLOPE OF A CUEVE 37

From this and 9 AVO may at once deduce the theorem :

If the derivative is positive, the curve runs up to the right.

If the derivative is negative, the curve runs down to the right. Jfthe derivative is zero, the tangent to the curve is parallel to OX.

If the derivative is infinite, the tangent to the curve is perpendic-ular to OX.

difThe values of x which make ~~ zero or infinite are of par-

dx

ticular interest m the plotting of a curve. If the derivative

changes its sign at such a point, the curve will change its cliiec-

tion from down to up or from up to down. Such a point will

be called a turning-point. If y is an algebraic polynomial, its

derivative cannot be infinite; so we shall be concerned in this

chapter only with turning-points for which

^ = 0.Ax

They are illustrated m the two following examples:

Ex. 1. Consider equation (1) of 12,

y = 5 x a;8.

Plere ^ = 5 - 2 a: = 2^-dx \2

Equating ~ to zero and solving, we have x = - as a possible turning-

point It is evident that when x<- , is positive, and when #>- ~ is2 dx 2 dx

negative Therefoie x = \ corresponds to a turning-point of the curve at

which the lattei changes its direction from up to down It may be called

a high point of the cmve

Ex. 2. Consider

Here ^ = |(*a - 2* - 3) = |(a:

-3) (a + 1).

(IX o o

Equatingf-~- to zero and solving, we have x = 1 and x = 3 as possible

dvturning-points From the factored form of -^, and reasoning as in 9,

we see that when y < 1,~ is positive ;

when 1< c < 8, ^r is negative ;

38 DIFFERENTIATION

when x > 3, is positive. Therefore both x 1 and x = 3 give turmng-dx

points,the former giving a high point, and the latter a low point-

Substituting these values of x in the equa-

tion of the curve, we find the high point to

be ( lj 4f) and the low point to be (3, |).

The graph is shown in Fig 11

It is to be noticed that the solu-

tions of the equation do not^ dx

always give turning-points as illus-

trated in the next example.

Ex. 3. Consider

Heredx= xz - 6 x + 9 = (x 3)

2

Solving =0, we have x = 3

;but since the derivative is a perfect

dx

square, it is never negative Therefore x = 3 does not give a turning-point,

although when x = 3 the tangent to the curve is

parallel to OX. The curve is shown in Fig. 12.

The equation of the tangent to a curve

at a point (a^, yx ) is easily written down

We let ( -^ ) represent the value of -^ at

VaaJ/! ,, . dx

the point (a^, yx ).Then m = f-^j, and, n I, I X

from (3), 14, the equation of the tangent is

Ex. 4. Find the equation of the tangent at (1, 1) to the curve

We have

and

dx

Therefore the equation of the tangent is

wljich reduces to 1 = 0.

SECOND DERIVATIVE

From (2), 14, it also follows that if $ is the angle which

the tangent at any point of a curve makes with OX, then

EXERCISES

Locate the turning-points, and then plot the following curves :

1. 2/= 3aa

-f-4a: + 4. 4.2/= a-

8 - 6 #a+ 9 a; + 3.

2. ?/=3 + 3a!-2a;a. 5.

2/=

(2o;8 + 3a:

2 -12a;- 20)

3.2/= ce

8 -3a:2-j-4. 6. ?/

= 2 + 9ce -f 3a;a x*

'

7. Find the equation of the tangent to the curve y 3 2x -f- a2

at the point for which x = 2

8. Find the equation of the tangent to the curve 2/=l+3a: as2

3a;8

at the point for which x = 1.

9. Find the points on the curve y = a8 + 3a:2

3a; + 1 at which

the tangents to the curve have the slope 6

10. Find the equations of the tangents to the curve

y = xs + 2x* - x + 2

which make an angle 135 with OX

11. Find the equations of the tangents to the curve y=a58-f-a;

a 2xwhich are perpendicular to the line 3: + 2y + 4 =

12. Find the angle of intersection of the tangents to the curve

y = oj8-}- as

2 2 at the points for which x = 1 and x = 1 respectively.

16. The second derivative. The derivative of the derivative

is called the second derivative and is indicated by the symbol

-=-(] or -, We have met an illustration of this in theax \dx/ ax*

case of the acceleration. We wish to see now what the second

derivative means for the graph.

Since ~ is equal to the slope of the graph, -we havedx

d*y d , , ^j& ^ -=- (slope).dor dx

40 DIFFERENTIATION

From this and 9 we have the following theorem :

If the second derivative is positive, the slope is increasing as x

increases ;and if the second derivative is negative, the slope

is de-

creasing as x increases.

We may accordingly use the second derivative to distinguish

between the high turning-points and the low turning-points of

a curve, as follows:

If, when x a,- = and -r is positive, it is evident that

/7 du-^ is increasing through zero ; hence, when x < a,

-~ is nega-dx d

<lr

tive, and when x > a, -j-is positive. The point for which x = a

Cu3s

is therefore a low turning-point, by 9.

d'ii d^ySimilarly, if, when x = a, -^-

and -~ is negative, it is

duevident that -& is decreasing through zero ; hence, when x < a,j CbX J

-Z- is positive, and when x > a, -?- is negative. The point fordx dx

which x = a is therefore a high turning-point of the curve, by 9.

These conclusions may be stated as follows:

If -j-and -=-* is positive at a point of a curve, that point

^ "^ "

rlai fj^itl

is a low point of the curve. If -J-= and -~ is negative at a

CliK ClX

point of a curve, that point is a high point of the curve.

In addition to the second derivative, we may also have third,

d?u d^tifourth, and higher derivatives indicated by the symbols -,'{> '^etc. These have no simple geometric meaning.

EXERCISES

Plot the following curves after determining tlieir high and low-

points by the use of -^ and -r :

dx dor

= 3a;8 x 2 3. y = 7 18x 3a-aH- 4,r8,

MAXIMA AND MINIMA 41

17. Maxima and minima. If f(a) is a value of /() which

is greater than the values obtained either by increasing or by

decreasing x by a small amount, /(ff) is called a maximum value

of f(p). If /() is a value of /(#) which is smaller than the

values of f(j) found either by increasing or by decreasing x bya small amount, /(a) is called a minimum value of /(a?).

It is evident that if we place

and make the graph of this equation, a maximum value of /(of)

occurs at a high point of the curve and a minimum value at a

low point. From the previous sections we have, accordingly,the following rule for finding maxima and minima:

To find the values of x which give maximum or minimum lvalues

of y, solve the equation

^ = 0.dx

If x = a is a root of this equation, it must bo tested to see

whether it gives a maximum or minimum, and which. We have

two tests :

TEST I. If the sign of-^

changes from + to as x increasesCv*&

through a, then & = a (jives a maximum value of y. If the sign of

-~-changes from to + as x increases through a, tJien x = a gives

(K&

a minimum value of y.

H'U d^iiTEST II. If x = a makes -^ = and -~~ netHttive. then # = #

dx do* , ,/yy / ff *aj

gives a maximum value of y, Jf x a mal(e -~ = and -~~.ty^fj cfj

positive, then x a gives a minimum value of y.

Either of these tests may be applied according to convenience,

It may be noticed that Test I always works, while Test II fails

d\ito give information if

---^= when x a. It is also frequently

CtJs

possible by the application of common sense to a problem to

determine whether the result is a maximum or minimum, and

neither of the formal tests need then be applied.

42 DIFFERENTIATION

Ex. 1. A rectangular box is to be formed by cutting a square from

each, corner of a rectangular piece of cardboard and bending the resulting

figure. The dimensions of the piece of cardboard being 20 in by 30 m.,

required the largest box which can be made.

Let x be the side of the square cut out. Then, if the cardboard is bent

along the dotted lines of Fig 13, the dimensions of the box are 30 - 2 x,

20 2 x, x. Let V be the volume of the box.

Then p = (20- 2 a:) (30

- 2 *) 8Q&3D

7V= 600 - 200 x + 12 a;

2. "j

dxfiy I

-.-------1

-Equating ___ to zero, we have

13

whence x = 25 5 = 3.9 or 12 7.

The result 12.7 is impossible, since that amount cannot be cut twice

fiom the side of 20 in The result 3.9 corresponds to a possible maximum,and the tests are to be applied.

dVTo apply Test I we write in the factored form

dx

dVwhen it appears that - changes from + to , as x increases through 3.9

ax

Hence x = 3.9 gives a maximum value of V.

dsVTo apply Test II we find - = - 200 + 24 x and substitute x = 3.9.

dyr

The result is negative. Therefore x = 3.9 gives a maximum value of V.

The maximum value of V is 1056 + cu. in., found by substitutingx =s 3.9 in the equation for V.

Ex. 2. A piece of wood is in the form of a right circular cone, the

altitude and the radius of the base of which are each equal to 12 in. Whatis the volume of the largest right circular cylinder that can be cut fromthis piece of wood, the axis of the cylinder to coincide with the axis of

the cone ?

Let x be the radius of the base of the required cylinder, y its altitude,and V its volume. Then

V= vx*y. (1)

We cannot, however, apply our method directly to this value of V, since

it involves two variables x and y. It is necessary to find a connection

MAXIMA AND MINIMA 43

between x and y and eliminate one of them. To do so, consider Fig. 14,

which is a cross section of cone and cylinder. From smiilai triangles we have

FE _AD t

EC DC'

that18'

whence y 12 x.

Substituting in (1), we have

whencedVdx

24 vx 8 irx*.

dVEquating to zero and solving, we find

dx Bx = or 8. The value x = is evidently not a

solution of the problem, but x = 8 is a possible

solution.

Applying Test I, we find that as x increases through the value 8,

dxchanges its sign from + to . Applying Test II, we find that

24 ir 6 irx is negative when x = 8. Either test shows that x = 8(IX

corresponds to a maximum value of V. To find V substitute x = 8 in the

expression for V. We have V = 256 ir cu. in.

EXERCISES

1. A piece of wire of length 20 in. is bent into a rectangle one

side of which is a:. Find the maximum, area.

2. A gardener has a certain length of wire fencing with which to

fence three sides of a rectangular plot of land, the fourth side beingmade by a wall already constructed. Required the dimensions of

the plot which contains the maximum area.

3. A gardener is to lay out a flower bed in the form of a sector

of a circle. If he has 20 ft. of wire with which to inclose it, whatradius will he take for the circle to have his garden as large as

possible ?

4. In a given isosceles triangle of base 20 and altitude 10 a rec-

tangle of base x is inscribed. Find the rectangle of maximum area.

6. A right circular cylinder with altitude 2 as is inscribed in a

sphere of radius a. Find the cylinder of maximum, volume,

44 DIFFEEENTIATION

6. A rectangular box with a square base and open at the top is

to be made out of a given amount of mateiial. If no allowance is

made for the thickness of the material or for waste in construction,

what are the dimensions of the largest box that can be made ?

7. A piece of wire 12 ft. in length is cut into six portions, two

of one length and four of another Each of the two former portions

is bent into the form of a square, and the corners of the two squares

are fastened together by the remaining portions of wire, so that the.

completed figure is a rectangular parallelepiped Find the lengths

into which the wire must be divided so as to produce a figure of

maximum volume

8. The strength of a rectangular beam, varies as the product oE

its breadth and the squaie of its depth Find the dimensions of the

strongest rectangular beam that can be cut from a circular cylindri-

cal log of radius a inches

9. An isosceles triangle of constant perimeter is revolved about

its base to form a solid of revolution. What are the altitude andthe base of the triangle when the volume of the solid generated is

a maximum ?

10 The combined length and girth of a postal parcel is 60m.Find the maximum volume (1) when the parcel is rectangular with

square cross section, (2) when it is cylindrical

11. A piece of galvanized iron 5 feet long and a feet wide is to be

bent into a U-shaped water drain I feet long. If we assume that the

cross section of the drain is exactly represented by a rectangle on

top of a semicircle, what must be the dimensions of the rectangleand the semicircle in order that the drain may have the greatest

capacity (1) when the drain is closed on top9

(2) when it is openon top ?

12. A circular filter paper 10m. in diameter is folded into a

right circular cone. Find the height of the cone when it has the

greatest volume

18. Integration. It is often desirable to reverse the process of

differentiation. For example, if the velocity or the acceleration

of a moving body is given, we jftay wish to find the distance

traversed ; or if the slope of a curve is given, we may wish to

find the curve.

INTEGKATION 45

The inverse operation to differentiation is called integration,

and the result of the operation is called an integral. In the case

of a polynomial it may be performed by simply working the

formulas of differentiation backwards. Thus, if n is a positive

integer and ,

then

The first term of this formula is justified by the fact that if it

is differentiated, the result is exactly aa?. The second term is

justified by the fact that the derivative of a constant is zero.

The constant may have any value whatever and cannot be

determined by the process of integration. It is called the constant

of integration and can only be determined in a given problem by

special information given in the problem. The examples will

show how this is to be done.

Again, if

^ =,

dx

then ;/= ax + C. (2)

This is only a special case of (1) with n = 0.

Finally, if

= aQrn+ a^" l+ + an^x + an,

ax

C8)

Ex. 1. The velocity v with which a body is moving along a straight

line AB (Fig. 15) is given by the equation

How far will the body move in the J'IG. 15

time from * = 2 to * = 4?

If when t = 2 the body is at Pv and if when t = 4 it is at PB , we are

to find P,PS .

(IV

By hypothesis, -j= 10 1 + 5.

Therefore s- B i* + 5 1 + C. (1)

46 DIFFERENTIATION

have first to determine C. If s la measured from Pv it follows that

when t_

2> s - o.

Therefore, substituting in (1), we have

= 8(2)2 + 5 (2) + C;

whence C=-42,

and (1) becomes s = 8 i3 + 5 1 - 42. (2)

This is the distance of the body from Plat any time /. Accordingly, il>

remains for us to substitute t = 4 m (2) to find the loquiied distance 1\I\,

Thei e results _ 2 _ 42 _ 106>

If the velocity is in feet per second, the required distance is in foot.

Ex. 2. Required the curve the slope of which

at any point is twice the abscissa of the point.

By hypothesis, -j-= 2 x

CIX

Therefore y = JB* + C (1)

Any curve whose equation can be derived

fiom (1) by giving C a definite value satisfies

the condition of the problem (Fig. 16) If it

is required that the curve should pass throughthe point (2, 3), we have, from (1),

3 = 4 + <?;

whence C = 1,

and therefore the equation of the curve is

PICK 10

But if it is required that the curve should pass through ( 8, 10),

we have, from (1), 10 = 9 + C;

whence C = 1,

and the equation is y = 2 + 1

EXERCISES

In the following problems v is the velocity, in feet per second, of

a moving body at any time t

l. If v = 32 1 + 30, how far will the body move in the time from

AEEA 47

2. If v = 3 1* + 4 1 + 2, how far will the body move in the time

from t = 1 to t = 3 f

3. If v = 20 + 25, how far will the body move in the fourth

second ?

4. If v = t2 2 1 + 4, how far will the body move in the fifth

and sixth seconds ?

5. If v 192 32 1, how far will the body move before v 9

6. A curve passes through the point (1, 1), and its slope at

any point (x,y) is 3 more than twice the abscissa of the point.What is its equation

?

7. The slope of a curve at any point (aj, y) is Go;2 + 2x 4, andthe curve passes through the point (0, 6) What is its equation ?

8. The slope of a curve at any point (x, y) is 4 3# jc2,and

the curve passes through the point ( 6, 1). What is its equation ?

9. A curve passes through the point (5, 2), and its slope at anypoint (05, y) is one half the abscissa of the point What is its equation ?

10. A curve passes through the point ( 2, 4), and its slope at

any point (x, y) is a;2 x + 1. What is its equation ?

19. Area. An important application of integration occurs in

the problem of finding an area bounded as follows :

Let US (Fig. 17) be any curve with the equation y =/(&), and

leitkED and BC be any two ordinates. It is required to find the

area bounded by the curveRS, ythe two ordinates J3D and J3C,

and the axis of x.

Take -MP, any variable or-

dinate between MD and J3<7,

and let us denote by A the

area EMPD bounded by the

curve, the axis of #, the fixed

ordinate J8I>, and the variable Q jj jf jy" B -^

ordinate MP.jUlG J7

It is evident that as values

are assigned to a; =OJf, different positions of MP and correspond-

ing values of A are determined. Hence A is a function of x for

dAwhich we will find -r

ax

48 DIFFERENTIATION

Take MN=Ax and draw the corresponding ordinate NQ.

Then the area MNQP=&A. If L is the length of the longest

ordinate of the curve between MP and NQ, and s is the length

of the shortest orduiate in the same region, it is evident that

sA# < Au4 < LAa;,

for L&x is the area of a rectangle entirely surrounding AJ, and

is the area of a rectangle entirely included m A^t.

Dividing by A&, we have

. A.1 rs < < i.Aa;

As Aa; approaches zero, JVQ approaches coincidence with MP,and hence * and L, which are always between NQ and MP,

approach coincidence with MP. Hence at the limit we have

~ =MP^ y =/(*). (1)

Therefore, by integrating,

A=F(x) + C, (2)

where F(x) is used simply as a symbol for any function whose

derivative is /(#)We must now find C. Let OE= a. When MP coincides with

ED, the area is zero. That is, when

x = a, A = 0.

Substituting in (2), we have

whence C= F(a),and therefore (2) becomes

^=^(00-^(0). (8)

Finally, let us obtain the required area JSBCD. If OBft,

thin

will be obtained by placing x = b in (3). Therefore we have,

finally>JFta). (4)

AEEA 49

In solving problems the student is advised to begin with

formula (1) and follow the method of the text, as shown in

the following example:

Ex. Find the area bounded by tho axis of x, the curve y = ^xz, and the

ordmates x = 1 and x = 3.

In Fig 18, BE is the line x = l, CD is the line x = 3, and the requiredarea is the a.re&J3CDE Then, by (1),

~dx=

SX*'

whence A ~ \ x8 + C

' When a;= 1, A = 0, and therefore

whence C =-J,

and A =J a:

3%

Finally, when^

x s=3,

EXERCISES

1. Find the area bounded by the curve y = 4 a1a;2,the axis of

x, and the lines x = 1 and a: = 3.

2. Find the area bounded by the curve y = s* + 8 a; + 18, the

axis of x, and the lines x = 6 and x = 2.

3. Find the area bounded by the curve y = 1C + 12 x xs,the

axis of x, and the lines x = 1 and a; = 2

4. Find the area bounded by the curve y + a;2 9 = and the

axis of x. .

5. Find the area bounded by the axis of x and the curve

y = 2 x a;2

.

6. Find the smaller of the two areas bounded by the curve

y = 5 x* xa,the axis of x, and the line x = 1.

7. Find the area bounded by the axis of x, the axis of y, and the

curve Ayx3605 + 9.

8. Find the area bounded by the curve y = x*~ 2 3<taj + 8

and the axis of x.

50 DIFFERENTIATION

9. If A denotes the area bounded by the axis of y, the curve

a; =/(y), a fixed line y ~ b, and any variable line parallel to OX,

prove that7

10. If A denotes an area bounded above by the curve //=/(*")>

below by the curve y = F(x), at the left by the fixed line x = (t,

and at the right by a variable ordinate, prove that

20. Differentials. The derivative has been denned as the limit

of and has been denoted by the symbol ~. This symbolAa? clx

is in the fractional form to suggest that it is the limit of a

fraction, but thus far we have made no attempt to treat it as

a fraction.

It is, however, desirable in many cases to treat the derivative

as a fraction and to consider dx and dy as separate quantities.

To do this it is necessary to define dx and dy in such a mannerthat their quotient shall be the derivative. We shall begin by

defining dx, when x is the independent variable; that is, the

variable whose values can be assumed independently of any oilier

quantity.

We shall call dxth& differential of x and define it as a changein x which may have any magnitude, but which is generally

regarded as small and may be made to approach aero as a

limit. In other words, the differential of the independent variable

x is identical with the increment of x ; that is,

dx sss A*. (1)

After dx has been defined, it is necessary to define dy so

that its quotient by dx is the derivative. Therefore, if y =/(V)

and -j- =/'(;), we havedx

dyt

That is, the differential of the function y is equal to the derivative

times the differential of the independent variable x.

DIFFERENTIALS 51

In equation (2) the derivative appears as the coefficient of dx.

For this reason it is sometimes called the diff&renticd coefficient.

It is important to notice the distinction between dy and Ay.The differential dy is not the limit of the increment A#, since both

dy and ky have the same limit, zero. Neither is Ay equal to a verysmall increment A#, since it generally differs in value from Ay.It is true, however, that when dy and Ay both become small, theydiffer by a quantity which is small compared with each of them.

These statements may best be under-

stood from the following examples :

Ex. 1. Let A be the area of a square with

the side x so that

If a; is increased by Ao; = dx, A is increased

by A.4, where

A4 =(a? + dxf

- x* = 2 x dx + (dx)a. & ^

Now, by (2), dA = 2zdx, FNJ. ig

so that A.I and dA differ by (dx)'2

Referring to Fig 19,.we see that dA is represented by the rectangles (1)

and (2), while A.4 is represented by the rectangles (1) and (2) togetherwith the square (3) , and it is obvious from the figuie that the square (8)

is very small compared with the rectangles (1) and (2), provided djc is

taken small. For example, if x = 5 and dx = .001, the rectangles (1)

and (2) have together the area 2 z die = .01 and the square (3) has the

area .000001.

Ex. 2. Let s - 16 ts,

where $ is the distance traversed by a moving body in the time t

If t is increased by Al = dt, we have

As = 10 (t + dt)*- 1C & - 32 tdi + 16 (dt}\

and, from (2), ds = 32 1 dt;

so that As and ds differ by 16 (dt)z The terra 16

(rf/)a is very small com-

pared with the term BZtdt, if dt is small. For example, if f = 4 and

dt - .001, then 32 tdt = ,128, while 16(eft)

2 .000016

In this problem As is the actual distance traversed in the time dt, andda is the distance which would have been traversed if the body had moved

throughout the time dt with the same velocity which it had at the begin-

ning of the time dt.

52 DIFFERENTIATION

In general, if y=f(%) and we make a graphical representa-

tion, we may have two cases as shown in Figs. 20 and 21.

In each figure, WN=PR &v = dx and RQ=ky, since HQis the total change in y caused by a change of dx =M.N in x.

If PT is the tangent to the curve at P, then, by 15,

so that, by (2), dy = (tan HPT) (P7?) = R T.

R

M N C M N

FIG. 20 PIG 21

In Fig. 20, dy < A#, and in Fig. 21 dy >&y; but in each case

the difference between dy and Ay is represented m magnitudeby the length of QT.

This shows that RQ =Ay is the change in y as the point JP is

supposed to move along the curve /=/(^') while RT dy is

the change in the value of y as the point P is supposed to movealong the tangent to that curve. Now, as a very small arc docs

not deviate much from its tangent, it is not hard to see graphi-

cally that if the point Q is taken close to P, the difference between

RQ and RT, namely, QT, is very small compared with RT.A more rigorous examination of the difference between the

increment and the differential lies outside the range of this book.

EXERCISES

1. If y = x6 - 3ic2 + ix + 1, find dy.

2. If y = x4 + 4 8 - x* + 6.-B, find dy.

3. If V is the volume of a cube of edge x, find both AF and dVand interpret geometrically.

APPROXIMATIONS 53

4. If A is the area of a circle of radius r, find both Avl and dA.

Show that A.4 is the exact area of a ring1 of width dr, and that dA

is the product of the inner circumference of the ring by its width.

5. If V is the volume of a sphere of radius r, find AF and dV.

Show that A V is the exact volume of a spherical shell of thickness

dr, and that dV is the product of the area of the inner surface of

the shell by its thickness.

6. If A is the area described in 19, show that dA = ydx Show

geometrically how this differs from A/i

7. If s is the distance traversed by a moving body, t the time,

and v the velocity, show that ds = vdt. How does ds differ from A& 9

8. If y = xz and x = 5, find the numerical difference between

dy and A?/, with successive assumptions of dx = 01, dx, = .001, and

dx = .0001.

9. If y = x8 and x = 3, find the numerical difference between dyand Ay for dx = .001 and for dx = 0001

10. For a circle of radius 4 in. compute the numerical difference

between dA and AJ. corresponding to an increase of r by .001 in.

11. Eor a sphere of radius 3 ft. find the numerical difference

between dV and A V when r is increased by 1 in.

21. Approximations. The previous section brings out the fact

that the differential of y differs from the increment of y by a

very small amount, which becomes less the smaller the incre-

ment of x is taken. The differential may be used, therefore, to

make certain approximate calculations, especially when the ques-tion is to determine the effect upon a function caused by small

changes in the independent variable. This is illustrated in the

following examples:

Ex. 1. Find approximately the change in the area of a square of side

2 in. caused by an increase of .002 in. in. the side.

Let x be the side of the square, A its area. Then

A = x2 and dA = 2 xdx.

Placing x 2 and dx =* .002, we find dA = .008, which is approximatelythe required change in the area.

If we wish to know how nearly correct the approximation is, we may com-

pute A4 = (2.002)8-

(2)2 = .008004, which is the exact change in A. Our

approximate change is therefore in error by .OOOQOAj a very email amount.

54 DIFFERENTIATION

Ex. 2. Find approximately the volume of a sphere of radius 1.9 in.

The volume of a sphere of radius 2 in. is -^ TT, and the volume of the

required sphere may be found by computing the change in the volume of

a sphere of radius 2 caused by decreasing its radius by .1.

If r is the radius of the sphere and V its volume, we have

F= Tjr8 and dV = 4 vr*dr.

Placing r = 2 and dr .1, we find dV= 1.6 TT. Hence the volume

of the required sphere is approximately

To find how much this is in error we may compute exactly the volume

of the required sphere by the formula

V= ^7r(1.9)8 = 9.1453 ir.

The approximate volume is therefore in error by .0786 TT, which is less

than 1 per cent of the true volume.

EXERCISES

1. The side of a square is measured as 3 ft. long. If this length

is in error by 1 in,find approximately the resulting error in the area

of the square.

2. The diameter of a spherical ball is measured as 2 in., and the

volume and the surface are computed. If an error of^ in. has boon

made in measuring the diameter, what is the approximate error in

the volume and the surface ?

3. The radius and the altitude of a right circular cone are meas-

ured as 3 in. and 5 in. respectively. What is the approximate error

in the volume if an error of ^ in. is made in the radius ? What is

the error in the volume if an error of ^ in. is made in the altitude ?

4. Find approximately the volume of a cube with 3.0002 in. on

each edge.

5. The altitude of a certain right circular cone is the same as

the radius of the base. Find approximately the volume of the cono

if the altitude is 3.00002 in.

6. The distance * of a moving body from a fixed point of its

path, at any time t, is given by the equation $ = 16 ta-f- 100 1 50.

Find approximately the distance when t s= 4.0004.

7. Find the approximate value of v?+ x 2 when x =s 1,0003.

8. Find approximately the value of as*+ *a + 4 when = .99989.

GENERAL EXERCISES 55

9. Show that the volume of a thin cylindrical shell is approxi-

mately equal to the area of its inner surface times its thickness.

10. If V is the volume and S the curved area of a right circular

cone with radius of its base r and its vertical angle 2 a, show that

V s= I 7n<s ctn a and S = irr1 esc a Thence show that the volume of

a thin conical shell is approximately equal to the area of its inner

surface multiplied by its thickness.

GENERAL EXERCISES

Find the derivatives of the following functions from the definition

3 + 2x 1 6. VJB>' '

*

2. ! 4.a x

8. Prove from the definition that the derivative of - is ,

1

C C

9. By expanding and differentiating, prove that the derivative

of (2 x + 5)8

is 6(2a; + 5)2.

10. By expanding and differentiating, prove that the derivative

of (a2+ 1)

8is 6 a; (a

2 + 1)2.

11. By expanding and differentiating, prove that the derivative

of(a; + a)

nis n(x + a)""

1}where % is a positive integer.

12. By expanding and differentiating, prove that the derivative

of(a;

2+ as

)n is 2 nx(x

i + a2)"-1,where n is a positive integer.

13. Find when x*+ 8 a;84- 24 a8+ 32 a- + 16 is increasing and

when decreasing, as x increases.

14. Find when 9 x* 24 of 8 x* -f- 32 x + 11 is increasing and

when decreasing, as x increases.

15. Find a general rule for the values of x for which ace2 + bx + e

is increasing or decreasing, as x increases.

16. Find a general rule for the values of x for which x* a?x + 1

is increasing or decreasing, as x increases.

17. A right circular cone of altitude x is inscribed in a sphere of

radius a. Find when an increase in the altitude of the cone will cause

an increase in its volume and when it will cause a decrease,

A-B* HINT, In these examples make use of the relation vA'vIt =

56 DIFFERENTIATION

18. A particle is moving in a straight line in such a manner that

its distance x from a fixed point A of the straight line, at any time t,

is given by the equation x ts 9 & + 15 1 + 100. When will the

particle be approaching A ?

19. The velocity of a certain moving body is given by the equationv = t

z 7 1 -f- 10. During what time will it be moving in a direction

opposite to that in which s is measured, and how far will it move ?

20. If a stone is thrown up from the surface of the earth with a

velocity of 200 ft. per second, the distance traversed in t seconds is

given by the equation s = 200 1 16 z5a. Find when the stone moves

up and when down

21. The velocity of a certain moving body, at any time t, is given

by the equation v = t2 8 1 + 12 Find when thu velocity of the

body is increasing and when decreasing.

22. At any time t,the distance s of a certain moving body from a

fixed point in its path is given by the equation s= 16 24 1 -f 9 tf

2f.

When is its velocity increasing and when decreasing ? When is its

speed increasing and when decreasing?

23. At any time t, the distance of a certain moving body from a

fixed point in its path is given by the equation s t6 6 1* -j- 9 1 + 1

When is its speed increasing and when decreasing ?

24. A sphere of ice is melting at such a rate that its volume is

decreasing at the rate of 10 cu in. per minute. At what rate is the

radius of the sphere decreasing when the sphere is 2 ft. in diameter ?

25. Water is running at the rate of 1 cu. ft. per second into a

basin in the form of a frustum of a right circular cone, the radii of

the upper and the lower base being 10 in and 6 in. respectively, andthe depth being 6 in. How fast is the water rising in the basin wluwit is at the depth of 3 in. ?

<. 26. A vessel is in the form of an inverted right circular cone thevertical angle of which is 60. The vessel is originally filled with

liquid which flows out at the bottom at the rate of 3 cu. in. per minutw.At what rate is the inner surface of the vessel being exposed whenthe liquid is at a depth of 1 ft. in the vessel ?

27. Find the equation of the straight line which passes throughthe point (4, 6) with the slope .

28. Find the equation of the straight line through the points

(2,-

3) and (- 3, 4).

GENERAL EXERCISES 57

29. Find the equation of the straight lino determined by the points

(2, -4) and (2, 4).

30. Find the equation of the straight line through the point,

(1, 3) and parallel to the line a? 2 y -f- 7 =31. Find the equation of the straight lino tli rough the point

(2, 7) and perpendicular to the lino 2 . -f 4 // + 9 =32. Find the angle between the straight lines 2,r -|- 3// + H =

and y + 3 x -f- 1 = 0.

Find the turning-points of the following curves and draw the

graphs

33. y=3 as-ai"

34.2/= 16ca -40a- + 25

35. y } (x2 - 4 x - 2).

36. y = a* - 6 a* - 15 x + 5.

37. y = 3-3 + 3 as* 9 a: 14.

38. Find the point of intersection of the tangents to the curve

y = x + 2 a;2 xs at the points for which y = 1 and a? = 2

respectively

39. Show that the equation of the tangent to the curve y = o"2

+ 2bx + G at the point (x1} y^) is y = 2 (ca^-f /;), a,r,a+ 0.

40. Show that the equation of the tangent to the curve y a*a

+ ax + fi at the point (a^, ya) is y = (3 a

a + )*" 2 , -|_ />.

41. Find the area of the triangle included between the coordintitc

axes and the tangent to the curve y = a;8 at the point (2, 8).

42. Find the angle between the tangents to the curve //=-2./'"

-f- 4 a;2 x at the points the abscissas of which arc 1 and 1

respectively.

43. Find the equation of the tangent to the curve y =s xn 3 a;9

-+ 4.x 12 which has the slope 1.

44. Find the points on the curve;//= 3 .r,

B 4 a*,2 at which the

tangents are parallel to the lino x y = 0.

V 1 46. A length I of wire is to be cut into two portions which aw to

be bent into the forms of a circle and a square respectively. Showthat the sum of the areas of those figures will bo least when the wireis cut in the ratio IT : 4.

58 DIFFERENTIATION

46. A log in the form of a -frustum of a cone is 10 ft. long, the

diameters of the bases being 4 ft. and 2 ft. A beam with a square

cross section is cut from it so that the axis of the beam eoiuuidos

with the axis of the log. Find the beam of greatest volume that

can be so cut

47. Required the right circular cone of greatest volume which can

be inscribed in a given sphere.

48. The total surface of a regular triangular prism is to be Jt. Find

its altitude and the side of its base when its volume is a maximum.

49. A piece of wire 9 in. long is cut into five pieces, two of ouo

length and three of another Each of the two equal pieces is bent

into an equilateral triangle, and the vertices of the two triangles are

connected by the remaining three pieces so as to form a regular

triangular prism. How is the wire cut when the prism has the largestvolume ?

50. If t is time in seconds, v the velocity of a moving body in

feet per second, and v = 200 32 1, how far will the body move in

the first 5 sec. ?

51. If v = 200 32 1, where v is the velocity of a moving bodyin feet per second and t is time in seconds, how far will the bodymove in the fifth second ?

62. A curve passes through the point (2, 8), and its slope at

any point is equal to 3 more than twice the abscissa of the point.Find the equation of the curve.

53. A curve passes through the point (0, 0), and its slope at anypoint is oj

a 2 x + 7. Find its equation.

54. Find the area bounded by the curve y + sa 16 = and llio

axis of ao,

5 5 . Find the area bounded by the curve y = 2 as8 15 a?

a4. 3G + 1

,

the ordinates through the turning-points of the curve, and OX.56. Find the area between the curve y xa and the straight lint?

y x -f- 6.

57. Find the area between the curves y =* jca and y = 18 a8

.

58. The curve y = ax* is known to pass through the point (h, Jc),Prove that the area bounded by the curve, the axis of

a-, and theline x = h is hk.

59. Compute the difference between &A and dA for the area 4 ofa circle of radius 5, corresponding to an increase of .01 in the radius.

GENERAL EXERCISES 59

60. Compute the difference between AF and dV for the volume Vof a sphere of radius 5, corresponding to an increase of 01 in the

radius

61. If a cubical shell is formed by increasing each edge of a cube

by dx, show that the volume of the shell is approximately equal to

its inside surface multiplied by its thickness.

62. If the diameter of a sphere is measured and found to be 2 ft,

and the volume is calculated, what is the approximate error in the

calculated volume if an error of in has been made in obtainingthe radius ?

63. A box in the form of a right circular cylinder is 6 in deepand 6 in. across the bottom. Find the approximate capacity of the

box when it is lined so as to be 5 9 in deep and 5 9 in across the

bottom.

64. A rough wooden model is in the form of a regular quadran-

gular pyramid 3 in. tall and 3 in. on each side of the base. After it

is smoothed down, its dimensions are all decreased by .01. What is

the approximate volume of the material removed ?

65. By use of the differential find approximately the area of a

circle of radius 1.99. What is the error made in this approximation ?

66. Find approximately the value of a;6+ 4 C

8+ a when x = 3.0002

and when x = 2.9998.

67. The edge of a cube is 2.0001 in. Find approximately its surface.

68. The motion of a certain body is defined by the equation

s = i8 + 3 i* + 9 1 27. Find approximately the distance traversed

in the interval of time from t = 3 to t = 3.0087.

CHAPTER III

SUMMATION

22. Area by summation. Let us consider the problem to find

the area bounded by the curve y ^ #2,the axis of r, and the

ordmates x = 2 and x 3 (Fig. 22). This may be solved by the

method of 19 ; but we wish to show that it may also be considered

as a problem in summation, since the area is approximately equalto the sum of a number of rectangles constructed as follows :

We divide the axis of x between x = 2 and x 3 into 10 parts,3 2

each of which we call A#, so that Ax = r =.1. If xl

is the

first point of division, xathe second point, and so on, and

rectangles are constructed as shown in the figure, then the

altitude of the first rectangle is (2)2, that of the second rec-

tangle is z 2

=|(2.1)2=.882, and so on. The area of tho

first rectangle is ^(2)2Ao;=.08, that of the second rectangle is

J 1

2Aa; = |(2.1)

2A=.0882, and so on.

Accordingly, we make the following calculation:

s = 2, -|(2)3Aa;= .08

0^=2.1, %(xy&x= .0882

#2=2.2, (z2)

2Az = .0968

z8=2.3, l(xy&x= .1058

z4 =2.4, %(xy&x= .1152

aja =2.5, O6)

2Aa= .1250

a;6=2.6, %(x^&x= .1352

z7=2.7, |(r 7)

2A^= .1458

z8 =2.8, (z8)

2Aa;= .1508

tf=2.9,fAr= -1G82

. ^i his is a first approximation to the area.

For a better approximation the axis of x between #= 2 and 05= 3

may be divided into 20 parts with Aa;= .05. The result is 1.241 8.

60

ABJEA 61

FIG. 22

If the base of the required figure is divided into 100 parts with

Az = .01, the sum of the areas of the 100 rectangles constructed

as above is 1.26167.

The larger the num-ber of parts into , which

the base of the figure is

divided, the more nearlyis the required area ob-

tained. In fact, the re-

quired area is the limit

approached as thenumberof parts is indefinitely

"~Q

increased and the size of

A approaches zero.

We shall now proceed to generalize the problem just handled.

Let LK (Fig. 23) be a curve with equation y =/(), and let

OEa and OB b. It is re-

quired to find the area bounded

by the curve LK, the axis of #,

and the ordmates at E and B.

For convenience we assume

in the first place that a < b

and that f(x) is positive for

all values of x between a and b.

We will divide the line EBinto n equal parts by placing

and laying off the

E MI Me Jf J/ Mj Mt B

FIG. 28

n

lengthsEM^-Let

_ 1J5=Aa: (in Fig. 23,n=9).

.., Jf^A* parallel to OX.

/(a)Aa;=:the area of the rectangle

/(#].)AOJ= the area of the rectangle

/(*2)As?= the area of the rectangle

_1), and

Thenalsc

the area of the rectangle

62 SUMMATION

The sum

/(a)A* +/(aOA* +/(aA* + . . . +/(^_ 1)Aa; (1)

is then the sum of the areas of these rectangles and equal to

the area of the polygon EDR^R^ . . . R^^^li^B. It is evident

that the limit of this sum as n is indefinitely increased is the

area bounded by ED, EB, BC, and the arc DC.

The sum (1) is expressed concisely by the notation

where S (sigma), the Greek form of the letter S, stands for the

word "sum," and the whole expression indicates that the sum

is to be taken of all terms obtained from/(,)Aa; by giving to i

in succession the values 0, 1, 2, 3, ., n 1, where x ~a.The limit of this sum is expressed by the symbol

Cb

Ja

where /is a modified form of S.

Hence / /(#)<fa;=Lim^/(a;f)A:c=:

i/a isOarea

It is evident that the result is not vitiated if ED or J3C is of

length zero.

23. The definite integral. We have seen in 19 that if A is

the area EBCD of 22,

where F(x) is any function whose derivative is/(). Comparingthis with the result of 22, we have the important formula

r h

Ja

The limit of the sum (1), 22, which is denoted by C /(*) dx,Jn

is called a definite integral, and the numbers a and b are called

DEFINITE INTEG-KAL 68

the lower limit and the upper limit*, respectively, of the definite

integral.

On the other hand, the symbol If(x)dx is called an indefinite

integral and indicates the process of integration as already de-

fined in 18.

Thus, from that section, we have

+ 0,

G,

and, in general, \ f(x)dx=F(x)+ C,

where F(x) is any function whose derivative is/(#)We may therefore express formula (1) in the following rule :

To find the value of \ f(x)dx, evaluate \f(x)dx, substitute

x = b and x: = a successively, and subtract the latter result fromthe former.

It is to be noticed that in evaluating I f(x)dx the constant

of integration is to be omitted, because if it is added, it dis-

appears in the subtraction, since

In practice it is convenient to express F^ F^a) by the

symbol [^()]> so that

Ex. 1. The example of 22 may now be completely solved. The requiredarea is

27 8 10

* The student should notice that the word "limit" is here used In a

quite different from that in which it is used when a variable is said to approacha limit (1).

64 SUMMATION

The expression /(a;) dx which appears in formula (1) is called

the element of integration. It is obviously equal to dF(x). In

fact, it follows at once from 19 that

dA=ydx=f(x) dx.

In the discussion of 22 we have assumed that y and dr are

positive, so that dA is positive. If y is negative that is, if the

curve m Fig. 23 is below the axis of x and if dx is positive,

the product ydx is negative and the yarea found by formula (1) has a nega-

tive sign. Finally, if the area required

is partly above the axis of x and

partly below, it is necessary to find

each part separately, as in the follow-

ing example:

Ex. 2. Fmd the aiea bounded by the

curve y = x3a;2 6 x and the axis of a-

Plotting the curve (Fig 24), we see that

it crosses the axis of x at the points

B (-2, 0), 0(0, 0), and C(3, 0). Hence

part of the area is above the axis of x and

part below. Accoidingly, we shall find it

necessary to solve the pioblem in two parts,

first finding the aiea above the axis of x

and then finding that below To find the

first area we proceed as in 22, dividing

the area up into elementary rectangles fox-

each of whichdA = ydx = (a.

8a;2 - 6 *) dx ,

whence A- C\x*- x*-Qx)dx = [I*4 -

jj3?

o

FIG. 24

Similarly, for the area below the axis of x we find, as before,

dA = tlx = xs xs

But in this case y = xsjcz G x is negative and hence dA is negative,

for we are making x vary fiom to 8, and therefore dx is positive, Tlioio-

foie we expect to find the result of the summation negative. In faot,

we have -sA -

I (Xs - x* - 6 x^dx = [| a:

4 ~J as* ~

J o

-^(3)-3(3)]-0 = -15i

DEFINITE INTEGRAL 65

.8

As we aie asked to compute the total aiea bounded by the ciuve audthe axis of x, we discard the negative sign in the last summation aud add

5J and 15 f, thus obtaining 21^s as the required result.

If we had computed the definite integral

i8 xz 6

a.) dx,

we should have obtained the icsult 10 fy, which is the algebraic sum of

the two portions of area computed separately.

Ex. 3. Find the area bounded by the two curves y = .r2 and y = 8 a;

2.

We diaw the curves (Fig 25)

?/ = ra f\ "\y x(.*-}

and y = 8 - a;8, (2)

and by solving their equations wefind that they mtcisect at the pointsP

x (2, 4)aud>3 (-2,4)The reqiuiod area OP^JIP^O is evi-

dently twice the area OJ\liO, since

both ciuves aie symmetiical with le-

spect to OY. Accoidmgly, wo shall

find the area OP^BO and multiply it

by 2. This lattei area may be found

by subtracting the aic>a ON^P^O fromthe area ON^P-^BO, each of these areas

being found as in the pievious example ;

or we may proceed as follows :

Divide ON^ into n parts dx, and

through the points of division draw

stiaight lines paiallol to OY, intersect-

ing both curves. Let one of those lines

be M^QlQ2 Through the]101nts Q:

and

(22 draw straight linos parallel to OXuntil they meet the next vortical line to the right, thcnoby formingthe rectangle Q^SQy, The aroa of such a rectangle may be takon as

dA and may be computed as follows: its base is ih % and ids altitude IH

Q^ = A/^ M^Q,! = (8 j 8) j"

u =8 2 .ca

;for M^Q^ is tho ordinato

of a point on the curve (2) and J/j^ the ordinate of a point ou (1).

Thcicfore , ,,

A

Finally, tho required area is 2(10) = 21 }

= C \R - 2 ra) rl,r =[83-- ?, 3*]

66 SUMMATION

EXERCISES

1. Find the area bounded by the curve 4 y as2 2 = 0, the axis

of x, and the lines x 2 and x = 2.

2. Find the area bounded by the curve y = a? 7 X* + 8 a + 16

the axis of x, and the lines x = 1 and as = 3

3. Find the area bounded by the curve y = 25 x 10 a;2

-f- oi8 and

the axis of a.

4. Find the area bounded by the axis of x and the curve

y = 25 - a2.

5. Find the area bounded by the curve y=4aj2 4aj -3 and

the axis of x.

6. Find the area bounded below by the axis of x and above bythe curve y = #8 4 a2 4 x + 16.

7. Find the area bounded by the curve y = 4 as8 Sec

2 9cc+18and the axis of #.

8. Find the area bounded by the curve x* + 2y 8=0 and tho

straight line x-)-2y 6 = 0.

9. Find the area bounded by the curve 3y aj2 = and the

straight line 2x + y 9 =10. Find the area of the crescent-shaped figure bounded by the

two curves y = x* + 7 and y = 2 xz-f- 3

11. Find the area bounded by the curves 4y=sa3a 4 a; and

a24o: + 42/ 24 = 0.

12. Find the area bounded by the curve o; + 3 =2/

a

2y andthe axis of y.

24. The general summation problem. The formula

F(a) (1)rfJa

has been obtained by the study of an area, but it may be givena much more general application. For if f(x) is any function

of x whatever, it may be graphically represented by the curve

y =/(). The rectangles of Fig. 23 are then the graphical rep-

resentations of the products f(z)dx, and the symbol f f(x)dxJa

GENERAL PROBLEM 6T

represents the limit of the sum of these products. We mayaccordingly say:

Any problem which requires the determination of the limit of the

sum of products of the type f(x) dx may be solved by the use of

formula (1).

Let us illustrate this by considering again the problem, alreadysolved in 18, of determining the distance traveled in the time

from t = ttto t =

3 by a body whose velocity v is known. Since

ds

"*#'we have ds = vdt,

which is approximately the distance traveled in a small interval

of time dt. Let the whole time from t =t^

to t = tzbe divided

into a number of intervals each equal to dt. Then the total dis-

tance traveled is equal to the sum of the distances traveled in

the several intervals dt, and hence is equal approximately to the

sum of the several terms vdt This approximation becomes better

as the size of the intervals dt becomes smaller and their number

larger, and we conclude that the limit of the sum of the terms vdt

is the actual distance traveled by the body. Hence we have, if s

is the total distance traveled,

a = I vdt.

AIf, now, we know v in terms of t, we may apply formula (1).

Ex. If v = 16 1 + 5, find the distance traveled in the time from t = 2 to

* = 4.

We have directly

s = f *(16 1 + 5) dt = [8 1* + 5*]= 106.

EXERCISES

1. At any time t the velocity of a moving body is 3 & 4- 2 1 ft. persecond. How far will it move in the first 6 sec.?

2. How far will the body in Ex. 1 move during the seventh second ?

3. At any time t the velocity of a moving body is 6+ 5 1 & ft. persecond, Show that this velocity is positive during the interval fromt= 1 to t f= 6, and find how far the body moves during that interval.

68 SUMMATION

4. At any time t the velocity of a moving body is 4)53 24 f +11 ft.

per second. During what interval of time is the velocity negative, andhow far will the body move during that interval ?

6. The number of foot pounds of work done in lifting a weight in

the product of the weight in pounds and the distance in feet throughwhich the weight is lifted. A cubic foot of water weighs 021 lb.

Compute the work done in emptying a cylindrical tank of depth 8 ft.

and radius 2 ft, considering it as the limit of the sum of the jnort'rt

of work done in lifting each thin layer of water to the top of the tank.

25. Pressure. It is shown m physics that the pressure onone side of a plane surface of area A, immersed in a liquid tit a,

uniform depth of h units below the Kin-face of the liquid, is equalto wJul, where w is the weight of a unit volume of the liquid.This may be remembered by noticing

that wliA is the weight of the column"

of the liquid which would be supported

by the area A.

Since the pressure is the same in

all directions, we can also determine ^IQ 2Qthe pressure on one side of a planesurface which is perpendicular to the surface of the liquid andhence is not at a uniform depth.

Let ABC (Fig. 26) represent such a surface and RS the line of

intersection of the plane of ABC with the surface of the liquid.Divide ABC into strips by drawing straight lines parallel to Jttf.

Call the area of one of these strips dA, as in 28, and the depth of

one edge h. Then, since the strip is narrow and horizontal, tho

depth of every point differs only alightly from 7i, and tho pres-sure on the strip is then approximately wMA, Talcing P m thototal pressure, we write ^p_ j

3*

The total pressure P is the sum of tho pressures on the several

strips and is therefore the limit of the sum of terms of thoform whdA, the limit being approached as the number of tho

strips is indefinitely increased and the width of each indefinitelydecreased. Therefore /

P= I wlidA,

PEESSUKE 69

where the limits of integration are to be taken so as to include

the whole area the pressure on which is to be determined. Toevaluate the integral it is necessary to express both h and dA111 terms of the same variable.

Ex. 1. Find the pleasure on one side of a rectangle BCDE (Fig 27),

wheie the sides BC and ED aiu each 4 ft. long, the sides BE and CD are

each 8 ft. long, immersed in watei so that the plane of the rectangle is

peipendiculai to the surface of the

water, and the side BC is paiallel &to the surface of the watei and 2 ft.

below at

In Fig. 27, LK is the line of inter-

section of the surface of the water

and the plane of the rectangle. Let M * Nbe the point of intersection of

LK and BE produced Then, if x is

measured downward from along E DBE, x has the value 2 at the point B ^and x has the value 5 at the point E.

We now divide BE into parts dx, and through the points of division

draw straight lines parallel to BC, thus dividing the given rectangle

into elementaly rectangles such as MNP.S

Therefore dA = area of MNRS = MN - MS - 4 dx.

Since MN is at a distance x below LK, the pressure on the elementary

rectangle MNRS is approximately wx(4: dx'). Accoidmgly, we have

and P = fG

4 nxdx = [2 vxffe* 2 w(6)- 2 w(2)

2 = 42 20

t/2^

For water, w - 02 J Ib = ^ T.

Hence we have finallyP = 2625 Ib. = 1 T% T

Ex. 2. The base CD (Fig. 28) of a triangle BCD is 7 ft., and its altitude

from B to CD is 5 ft This triangle is immersed in water with its plane

perpendicular to the suiface of the water and with CD parallel to the sur-

face,and 1 ft below it, B being below CD Find the total pressure on one

side of this triangle.

Let LK represent the line of intersection of the plane of the triangle

and the surface of the water. Then B is 6 ft below LK Let BX be per*

pendicular to LK and intersect CD at T We will measure distances

from B in the direction BX and denote them by x. Then, at the point B,

x has the value ,and at T, x has the value 5.

TO SUMMATION

Divide the distance BT into parts dx, and through the points of divi-

sion draw straight lines parallel to CD, and on ouch of these linen n

lower base construct a rectangle such as MNJR8, where J'l and F sire

two consecutive points of division %onBX. L K

Then BE = x,G\

EF = dx,

and, by similar triangles,

whence

CD BTMN x

and

Then dA = the area of MNRS =J xdx,

Since B is C ft. below LK, and BE = x, it follows that E is (fl-

a;) ft.

below LK.Hence the pressure on the rectangle is approximately

dP = (& a"?*) (6 or)

w = (-V1 war $ ?/JJK

S) rfj?,

/*"

/o

= (105 to - iJA w)- =

JL^- to - 2010^ Ib. = Ii4 T.

EXERCISES

s/ 1. A gate in the side of a dam is in the form of a square, 4 ft.

on a side, the upper side being parallel to and 1C It. bolow the surfaoo

of the water in the reservoir. What is the pressure on the gate '(

v* 2. Find the total pressure on one side of a triangle of base 6 ft.

and altitude 6 ft., submerged in water so that the altitude is vertical

and the vertex is m the surface of the water.

</ 3. Find the total pressure on one side of a triangle of base 4 ft.

and altitude 6 ft., submerged in water so that the base is horizontal,

the altitude vertical, and the vertex above the base and 4 ft, from tlio

surface of the water.

e base of an isosceles triangle is 8 ft. and the equal sidas

6 ft. The triangle is completely immersed in water, its baaeallel to and 6 ft. below the surface of the water, its alti-

g perpendicular to the surface of the water, and its vertex

uc.ijmg ,uove the base. Find the total pressure on one side of the

triangle.

VOLUME 71

v 6. Find the pressure on one side of an equilateral triangle, 6 It.

on a side, if it is partly submerged in water so that one vortex is

one foot above the surface of the water, the corresponding altitude

being perpendicular to the surface of the water.

\s. The gate in Ex. 1 is strengthened by a brace which runs

diagonally from one corner to another. Find the pressure on each

of the two portions of the gate one above, the other below, the

brace.

7. A dam is in the form of: a trapezoid, with its two horizontal

sides 300 and 100 ft. respectively, the longer side being at the top ;

and the height is 15 ft. What is the pressure on the dam when the

water is level with the top of the dam ?

8. What is the pressure on tho dam of Ex. 7 when the water

reaches halfway to the top of the dam ?

9. If it had been necessary to construct the dam of Ex. 7 with

the shorter side at tho top instead of the longer side, how much

greater pressure would the dam have had to sustain when the

reservoir is full of water ?

10. The center board of a yacht is in the form, of a trapezoid in

which the two parallel sides are 3 ft and 6 ft., respectively, in length,

and the side perpendicular to these two is 4 ft. in length. Assumingthat the last-named side is parallel to the surface of the water at

a depth of 2 ft., and that the parallel sides are vertical, find the

pressure on one side of the board.

11. Where shall a horizontal lino be drawn across tho gate of

Ex. 1 so that tho pressure on the portion above tho lino shall equalthe pressure on the portion below ?

26. Volume. The volume of a solid may be computed by di-

viding it into n elements of volume, dV^ and taking the limit*

of the sum of these elements as n is increased indefinitely, tho

magnitude of each element at the same time approaching aero.

The question in each case is the determination of the form of

the element dr. We shall discuss a comparatively simple case

of a solid such as is shown in Fig, 29.

In this figure let Olf be a straight line, and let tho distance

of any point of it from be denoted by h. At one end tho solid

is bounded Tby a plane perpendicular to OH at <7, where 00~ a>

72 SUMMATION

and at the other end it is bounded by a piano perpendicular to

OH at B, where OS = 5, so that it has parallel bases.

The solid is assumed to be such that the area A of any planesection made by a plane perpendicular to Oil at a point distant

h from can be expressed as a func-

tion of h.

To find the volume of such a solid

we divide the distance CB into n parts

dh, and through the points of division

pass planes perpendicular to OH. Wehave thus divided the solid into slices

of which the thickness is dh

Since A is the area of the base of a

slice, and since the volume of the slice

is approximately equal to the volumeof a right cylinder of the same base

and thickness, we write

dV=Adli.

The volume of the solid is then the limit of the sum of terms of

the above type, and therefore

/"'F=I

Adh.Jo.

It is clear that the above discussion is valid even when one

or both of the bases corresponding toh=a and 7t= fi, respectively,reduces to a point.

Ex.1. Let OY (Fig. 30)be an edge of a solid such

that all its sections made

by planes perpendicular to

OFaie rectangles, the sides

of a rectangle in a planedistant y fiorn being re-

spectively 2 y and yz We

shall find the volume in-

cluded between the planes j,IO go# = and?/ = 2 J.

Dividing the distance from # = 0toy = 2& into n parts <li/, and passingplanes perpendicular to OY, we form rectangles such as MffttS, wlioro, if

VOLUME 73

OM = y, MN - y* and MS = 2 y Hence the area MNRS = 2 ys, and the

volume of the elementary cylinder standing on MNRS as a base is

thatls'

Therefore V =

Ex. 2. The axes of two equal light circular cylinders of radius o inter-

sect at right angles. Required the volume common to the two cylinders.

Let OA and OB (Fig. 31) be the axes of the ^cylmdeis and OY the common perpendicular to

OA and 013 at then point of intersection 0. ThenOAD and OBD are quadrants of two equal chcles

cut from the two cylinders by the planes throughOY perpendicular to the axes OB and OA, and

OD = a Then the figure represents one eighthof the requned volume.

We divide the distance OD into n parts dy,

and through the points of division pass planes per-

pendicular to OY Any section, such as LMNP,is a square, of which one side NP is equal to

V OP Z ON* OP = a, being a radius of one of the cylinders, and hence,

as ON= y, _NP = Va2 - f

Accordingly, the area of LMNP = az yz,and the volume of the ele-

mentary cylinder standing on LMNP as a base is

whence V= f "(aa -

y^ dy - [a2//-

J y] = a8.

Hence the total volume is -1/ a8.t

This method of finding volumes is particularly useful whenthe sections of the solid made by parallel planes are bounded

by circles or by concentric circles. Such a solid may be gen-erated by the revolution of a plane area around an axis in its

plane, and is called a solid of revolution. We take the following

examples of solids of revolution :

Ex. 3. Find the volume of the solid generated by revolving about OXthe area bounded by the curve

?y

2 = 4 *, the axis of x, and the line x = 8.

The generating area is shown in Fig. 32, where AB is the line a- = 8.

Hence OA s= 3,

74 SUMMATION

Divide OA into n parts dx, and through the points of division pass

straight lines parallel to OY, meeting the curve. When the area is revolved

about OX, each of these lines, as MP, JVQ, etc., generates a circle, the planeof which is perpendicular to OX The area ^of the circle generated by MP, for example,is irMP z

, which is equal to iryz =

7r(4 x), if

Hence the area of any plane section of

the solid made by a plane perpendicular to

OX can be expressed in terms of its dis-

tance from 0, and we may apply the pieviousmethod for finding the volume.

Since the base of any elementary cylinderis 4 TTX and its altitude is dx, we have

HenceV=f*4

vxdx = [2 ** = 18 .

Ex. 4. Find the volume of the ring surface generated by revolving aboutthe axis of x the area bounded by the line y = 5 and the curve y = 9 x~

The line and the curve (Fig 33) are

seen to intersect at the points Pl (2, 5)^

and Pz (2, 5), and the ring is generated

by the area P^BP^P^ Since this area is

symmetrical with respect to OY, it is evi-

dent that the volume of the ring is twice

the volume generated by the area APZBA.

Accordingly, we shall find the latter volumeand multiply it by 2.

We divide the line OMZ= 2 (Mz being

the projection of P2 on OX) into n parts

dx, and through the points of division draw

straight lines parallel to OY and intersect-

ing the straight line and the curve Oneof these lines, asMQP, will, when revolved

about OX, generate a circular ring, theouter radius of which is MP = y = Q~x*and the inner radius of which isMQ *=.y

= 5

Hence the area of the ring is jjf M 'M

Accordingly,

= TT (56- 18 x9 + a?<).

FIG-

dV = TT (56- 18 w2 + a;*)

dx

S3

Accordingly, the volume of the ring is 2 (70$ w) = 140$ TT.

VOLUME 75

EXERCISES

1. The section of a certain solid made by any plane perpendicularto a given line Oil is a circle with one point in OH and its center

on a straight line OB intersecting Oil at an angle of 46 If the

height of this solid measured from along Oil is 4ft., find its

volume by integration.

2. A solid is such that any cross section perpendicular to an

axis is an equilateral triangle of which each side is equal to the

square of the distance of the plane of the triangle from a fixed pointon the axis. The total length of the axis from the fixed point is 5.

Find the volume.

3. Find the volume of the solid generated by revolving about OXthe area bounded by OX and the curve y = 4 as a2

.

4. Find the volume of the solid generated by revolving about OXthe area included between the axis of x and the curve y 2 3 x*

5. Find the volume of the solid generated by revolving about the

line y = 2 the area bounded by the axis of y, the lines x = 3

and y = 2, and the curve y = 3 a:2.

6. On a spherical ball of radius 5 in. two great circles are drawn

intersecting at right angles at the points A and B* The material of

the ball is then cut away so that the sections perpendicular to ABare squares with their vertices on the two great circles. Find the

volume left.

7. Find the volume generated by revolving about the line x = 2

the area bounded by the curve if 8a?, the axis of a?, and the

line x = 2.

8. Any plane section of a certain solid made by a plane perpen-dicular to OF is a square of which the center lies on OY and two

opposite vertices lie on the curve y = 4 as2. Find the volume of the

solid if the extreme distance along OYis 3.

9. Find the volume generated by revolving about OY the area

bounded by the curve y2 8 x and the line x = 2.

10. Find the volume of the solid generated by revolving about OXthe area bounded by the curves y = 6 x as

2 and y s= flj8 6 a? + 10.

11. The cross section of a certain solid made by any plane perpen-

dicular to OX is a square, the ends of one of whose sides are on the

curves 16 y = y? and 4 y = y? 12, Find the volume of this solid

between the points of intersection of the curves.

T6 SUMMATION

GENERAL EXERCISES

1. The velocity in feet per second of a moving body at anytime t is t* 4 1 + 4 Show that the body is always moving in

the direction in which s is measured, and find how far it will move

during the fifth second

2. The velocity in feet per second of a moving body at any time

t is t* 4 1. Show that after t = 4 the body will always move in

the direction in which s is measured, and find how far it will movein the time from t = 6 to t = 9.

3. At any time t the velocity in feet per second of a movingbody is t

z 6 1 + 5 How many feet will the body move in the

direction opposite to that in which s is measured?

4. At any time t the velocity in miles per hour of a moving bodyis t* 2 1 3. If the initial moment of time is 12 o'clock noon, howfar will the body move in the time from 11.30 A M. to 2 p M. ?

y 5 . Find the area bounded by the curves 9y= 4xz and 45 9 y= a:3.

6. Find the total area bounded by the curves ?/2 =4a; and

T/2 = 4 a8 4 ax.

''7. Find the total area bounded by the curve y = xB and the

straight line y = 4 x.

8. Find the total area bounded by the curve y = x (x 1) (03 3)and the straight line y = 4

(a; 1).

9. ABCD is a quadrilateral with A 90, J3 = 90, AB 5 ft.,

BC = 2 ft., AD 4 ft. It is completely immersed in water with ABin the surface and AD and BC perpendicular to the surface. Findthe pressure on one side

10. Prove that the pressure on one side of a rectangle completelysubmerged with its plane vertical is equal to the area of the rectanglemultiplied by the depth of its center and by w (consider only thecase in which one side of the rectangle is parallel to the surface).

11. Prove that the pressure on one side of a triangle completelysubmerged with its plane vertical is equal to its area multiplied bythe depth of its median point and by w (consider only the case inwhich one side of the triangle is parallel to the 1

surface).12. The end of a trough, full of water, is assumed to be in the

form of an equilateral triangle, with its vertex down and its planevertical. What is the effect upon the pressure on the end if thelevel of the water sinks halfway to the bottom?

GENERAL EXERCISES 77

13. A square 2 ft on a side is immersed in water, with one vertex

in the surface of the water and with the diagonal through that

vertex perpendicular to the surface of the water. How much greateris the pressure on the lower half of the square than that on the

upper half?

14. A board is symmetrical with respect to the line AJB, and is of

such a shape that the length of any line across the board perpendic-ular to AB is twice the cube of the distance of the line from A.

AD is 2 ft. long The board is totally submerged in water, AB being

perpendicular to the suiface of the water and A one foot below the

surface. Find the pressure on one side of the board.

15. Find the pressure on one side of an area the equations of whose

boundary hues are x = 0, y = 4, and v/2 = 4 x respectively, where the

axis of x is taken in the surface of the water and where the positive

direction of the y axis is downward and vertical.

16. Find the volume generated by revolving about OX the area

bounded by OX and the curve 4 y = 16 a;2.

I,- 17. Find the volume generated by revolving about OX the area

bounded by the curve y = a2 + 2 and the line y = 3.

18. Find the volume generated by revolving about OX the area

bounded by OX and the curve y = 3 x x8.

19. Find the volume generated by revolving about the line y= 1

the area bounded by the curves 9 y = 2 x* and 9 y = 36 2 a;2

.

20. An axman makes a wedge-shaped cut in the trunk of a tree.

Assuming that the trunk is a right circular cylinder of radius 8 in.,

that the lower surface of the cut is a horizontal plane, and that the

upper surface is a plane inclined at an angle of 45 to the horizontal

and intersecting the lower surface of the cut in a diameter, find the

amount of wood cut out.

21. On a system of parallel chords of a circle of radius 2 there

are constructed equilateral triangles with their planes perpendicularto the plane of the circle and on the same side of that plane, thus

forming a solid. Find the volume of the solid.

22. Show that the volume of the solid generated by revolvingabout OY the area bounded by OX and the curve y = a Ix* is

equal to the area of the base of the solid multiplied by half its altitude.

23. In a sphere of radius a find the volume of a segment of one

base and altitude /*.

78 SUMMATION

24. A solid is sucli that any cross section perpendicular to an axis

is a circle, with its radius equal to the square root of the distance of

the section from a fixed point of the axis. The total length of the

axis from the fixed point is 4. Find the volume of the solid.

25. A variable square moves with its plane perpendicular to the

axis of y and with the ends of one of its diagonals respectively in

the parts of the curves y*= 16 x and ^ = 4 x, which are above

the axis of x Find the volume generated by the square as its planemoves a distance 8 from the origin.

26. The plane of a variable circle moves so as to be perpendicularto OX, and the ends of a diameter are on the curves y = a;

2 and

y = 3 cc2 8 Find the volume of the solid generated as the plane

moves from one point of intersection of the curves to the other.

27. All sections of a certain solid made by planes perpendicularto OF are isosceles triangles. The base of each triangle is a line

drawn perpendicular to OY, with its ends in the curve y = 4 a:2. The

altitude of each triangle is equal to its base Find the volume of

the solid included between the planes for which y = and y 6.

28. All sections of a certain solid made by planes perpendicularto Y are right isosceles triangles One leg of each triangle coincides

with the line perpendicular to OY with its ends in OY and the curve

y* 4 x. Find the volume of the solid between the sections for which

y = and y = 8.

29. Find the work done in pumping all the water from a full

cylindrical tank, of height 15 ft. and radius 3 ft., to a height of

20 ft. above the top of the tank.

30. Find the work done in emptying of water a full conical

receiver of altitude 6ft and radius 3ft., the vertex of the cone

being down.

CHAPTER IV

ALGEBRAIC FUNCTIONS

27. Distance between two points. Let P1 (a^, /x ) and Pz (rz , /a )

(Fig. 34) be any two points in the plane XOY, such that the

straight line P^PZ is not parallel either to OX or to OY. Through

P! draw a straight line parallel

to OJT, and through P2 draw a

straight line parallel to OY, and

denote their point of intersection

by R.

iPlTATl T^ 7? Al1* "-" 1" - tyJL UOll JL^J.V

* A-l*t/t'g

" "

*t*j

and

In the right triangle P^RPZ

whence

If

C1)

is parallel to 0JT, and the formula reduces to

J?J= JBa- ! (2)

In like manner, if ^-x^ P^ is parallel to OF, and the

formula reduces to

28. Circle. Since a ciVc?e is the locus of a point which is

always at a constant distance from a fixed point, formula (1)

27, enables us to write down immediately the equation of P

circle.

Let C(h, 7c) (Fig. 35) be the center of a circle of radius r.

Then, if P(x, y) is any point of the circle, by (1), 27, a? and

y must satisfy the equation

-^a=^ (1)79

80 ALGEBKAIC FUNCTIONS

Moreover, any point the coordinates of which satisfy (1),must be at the distance r from C and hence be a point of the

circle. Accordingly, (1) is the equation of a circle.

If (1) is expanded, it becomes

!

-2Aa;-2^ + A2+F r2=0, (2)

an equation of the second degree with no term in xy andwith the coefficients of a;

2 and /a

equal.

Conversely, any equation of the

second degree with no xy term and

with the coefficients of a;2 and y*

equal (as

Owhere A, G, F, and are any con- ^1Q g5

stants) may be transformed into

the form (1) and represents a circle, unless the number cor-

responding to rzis negative (see Ex. 3, page 81), in which

case the equation is satisfied by no real values of x and y and

accordingly has no corresponding locus.

The circle is most readily drawn by making such transfor-

mation, locating the center, and constructing the circle with

compasses.

Ex.1.

This equation may be written in the form

(x*-2x ) + (y2-4y )= 0,

and the terms in the parentheses may be made perfect squares by adding1 in the first parenthesis and 4 in the second parenthesis As we haveadded a total of 5 to the left-hand side of the equation, we must add an

equal amount to the right-hand side of the equation The result is

Cra - 2a: + 1) + (f - 4?/ + 4)

= 5,

which may be placed in the form

(-l) + <y-a)-5,the equation of a circle of radius V5 with its center at the point (1, 2).

CIRCLE 81

Ex. 2. 9 a:2 + 9 f - 9 x + 6 y

- 8 = 0.

Placing 8 on the right-hand side of the equation and then dividing by

9, we have** + ,*-*+ 3,

=|,

which may be treated by the method used in Ex. 1. The result is

the equation of a circle of ladius JV5, with its center at (\, i).

Ex.3. 9a;2 + 9y2-6a;-|-12y + ll =

Proceeding as in Ex. 2, we have, as the transformed equation,

an equation which cannot be satisfied by any real values of js and y, since

the sum of two positive quantities cannot be negative Hence this equation

corresponds to no real curve.

EXERCISES

1. Find the equation of the circle with the center (4, 2) and

the radius 3

2. Find the equation of the circle with the center (0, 1) and

the radius 5.

3. Find the center and the radius of the circle

9 = 0.

4. Find the center and the radius of the circle

- 6v - 15 = 0.

5. Find the equation of the straight line passing through the

center of the circle

and perpendicular to the line

2aj + 3v/ 4 =

6. Prove that two circles are concentric if their equations differ

only in the absolute term

29. Parabola. The locus of a point equally distantfrom a fixed

point and a fixed straight line is catted a parabola. The fixed

point is called the focus and the fixed straight line is called the

directrix,

82 ALGEBRAIC FUNCTIONS

Let F (Fig. 36) be tho focus and A\V the directrix of a

parabola. Through F draw a straight line perpendicular to

ftS, intersecting it at JP, and let this lino be the axis of jr.

Let the middle point of T>F be taken as 0, tho origin of coordi-

nates, and draw the axis QY. Thon, if tho distance, PA* is 2 <, Urn

coordinates of F are (0, 0) and tho equation of H$ is sr -~ - r.

Let /^(.r, y) be any point of tho parabola, and draw Uu

straight line FP and tho straight lino NP gy

perpendicular to US.

Then NP = + ,

and, by 27, FP- V(' <02+ //* /J

whence, from tho definition of tho parabola,

(a-*)** if- Or +<0"

which reduces to,?/

2= 4 6'a;. (1 )**

px(li ,w

Conversely, if the coordinates of any point /* satisfy (1), it

can be shown that the distances FP and NP aru equal, and

hence P is a point of the parabola.

Solving (1) for y in terms of a-, we havo

y = 2VS, (2)

We assume that e is positive. Thon it is evident that if a

negative value is assigned to #, y is imaginary, and no correspond-

ing points of the parabola can bo located. All poKitivo vahu'M

may be assigned to JR, however, and hence tho parabola IICH

entirely on the positive sidtj of tho axis OF.

Accordingly, wo assign positive valuos to #, compute tho nor-

responding values of y, and draw a smooth curve through tho

points thus located.

It is to be noticed that to every value tWHigncd to w lhro aro

two corresponding valuos of ?/, equal in magnitude and oppositein algebraic sign, to which there correspond two points of tho

parabola on opposite sides of OX and equally dintant from it.

Hence the parabola is ttymmMoal with rospect to CUT, and ac-

cordingly OX is called the gunk of the parabola.The point at which its axis intersects a parabola is called thew

tew of the parabola. Accordingly, is the vertex of the parabola.

PARABOLA 83

Returning to Fig. 36, if F is taken at the left of with the

coordinates ( c, 0), and RS is taken at the right of with the

equation x = c, equation (1) becomes

f = -kcx (3)

and represents a parabola lying on the negative side of OY.

Hence we conclude that any equation in the form

f=fa, (4)

where k is a positive or a negative constant, is a parabola, with

(kZ'and its directrix the straight line x'-^*

Similarly, the equation x*= Icy (5)

represents a parabola, with its vertex at and with its axis coin-

ciding with the positive or the negative part of OY, accordingas & is positive or negative. The focus is always the point

0,-^ j

and the directrix is the line y = - whether k be positive^"Y "i

or negative.

30. Parabolic segment. An important property of the parabolais contained in the following theorem :

The square of any two chords of a parabola which are perpen-dicular to its axis are to each other as their distances from the

vertex of the parabola.

This theorem may be proved as follows :

Let %(%!, #j.) and Pz (x^ 7/3 ) be any two points of any parab-ola /= Tex (Fig. 37).

Then y*= lex^

and yl 7ca ;

y\ X*whence , ~

yl **

whence

84 ALGEBRAIC FUNCTIONS

From the symmetry of the parabola, 2#1=

But a? = OM^ and x,== OMZ , and hence

(1) becomes

and 2#2=

FIG. 37

and the theorem is proved.

The figure bounded by the parabolaand a chord perpendicular to the axis

of the parabola, as Q^OP^ (Fig. 37),

is called a parabolic segment. The

chord is called the lase of the segment, the vertex of the

parabola is called the vertex of the segment, and the distance

from, the vertex to the base is called the altitude of the segment.

EXERCISES

Plot the following parabolas, determining the focus of each :

1. !/*= 8x 3. 2/

2 =:6a5.

2. cc2=42/. 4. aj

2= -7y.5. The altitude of a parabolic segment is 10 ft., and the length of

its base is 16 ft. A straight line drawn across the segment perpen-dicular to its axis is 10 ft. long. How far is it from the vertex of

the segment ?

6. An arch in the form of a parabolic curve, the axis being

vertical, is 50 ft. across the bottom, and the highest point is 15 ft.

above the horizontal. What is the length of a beam placed horizon-

tally across the arch 6 ft. from the top ?

7. The cable of a suspension bridge hangs in the form of a

parabola. The roadway, which is horizontal and 400ft. long, is

supported by vertical wires attached to the cable, the longest wire

being 80 ft. and the shortest being 20 ft. Find the length o a

supporting wire attached to the roadway 75 ft. from tho middle.

8. Any section of a given parabolic mirror made by a plane

passing through the axis of the mirror is a parabolic segment of

which the altitude is 6 in. and the length of the base 10 in. Findthe circumference of the section of the mirror made by a piano

perpendicular to its axis and 4 in. from, its vertex

ELLIPSE 85

9. "Find the equation of the parabola having the line x = 3 as its

directrix and having its focus at the origin of cooidmates.

10. Find the equation of the parabola having the line y 2 as

its directrix and having its locus at the point (2, 4).

31. Ellipse. The locus of a point the sum of whose distances

from two fixed points is constant is called an ellipse. The twofixed points are called the foci.

Let F and F' (Fig. 38) be the two loci, and let the distance

F'F be 2 c. Let the straight line determined by F' and F betaken as the axis of x, and the

middle point of F'F be taken

as 0, the origin of coordinates,

and draw the axis OY. Thenthe coordinates of F' and Fare respectively ( c, 0) and

(*, 0).

Let P(x, y) be anj' pointof the ellipse, and 2 a repre- n'

sent the constant sum of its 00Ju ICr uO

distances from the foci. Then,from the definition of the ellipse, the sum of the distances F'Pand FP is 2 a, and from the triangle F'PF it is evident that

2 a > 2 G; whence a > c.

By 27,

and FP=

whence, from the definition of the ellipse,

V(s-M)2+#2W(a-c) a+/= 2 - (1)

Clearing (1) of radicals, we have

O2- ca>2+ ay= a4- aV. (2)

Dividing (2) by a* aV, we have

86 ALGEBRAIC FUNCTIONS

But since ><?, a* c* is a positive quantity which may be

denoted by J3,and (3) becomes

Conversely, if the coordinates of any point P satisfy (4), it

can be shown that the sum of the distances Jf'P and FP ia 2 a,

and hence P is a point of the ellipse.

Solving (4) for y in terms of #, we have

y=s >/-** (5)

It is evident that the only values which can be assigned to x

must be numerically less than a; for if any numerically larger

values are assigned to r, the corresponding values of y are

imaginary, and no corresponding points can be plotted. Hencethe curve lies entirely between the lines x = a and x = a.

We may, then, assign the possible values to #, compute the

corresponding values of y, and, locating the corresponding points,

draw a smooth curve through them. As in the case of the pa-

rabola, we observe that OX is an axis of symmetry of the ellipse.

We may also solve (4) for x in terms of y, with the result

as = 5^-^. (0)

From this form of the equation we find that the ellipse lies

entirely between the lines ,y= --iandy = 5 and is symmetrical

with respect to OY.

Hence the ellipse has two axes, A'A and B'JB (Fig. 88), which

are at right angles to each other. But A1A= 2 a and JR'13 = 2 1) ;

and since a > 5, it follows that A'A > B'B, Hence A'A is called

the major axis of the ellipse, and fl'JJ is called the minor axis

of the ellipse.

The ends of the major axis, A' and A, are called the wrtwes of

the ellipse, and the point midway between the vertices is called

the center of the ellipse ; that is, is the center of the ellipse,and it can be readily shown that any chord of the ellipse which

passes through is bisected by that point.

ELLIPSE 87

From the definition of/>, c=* Vaa

#*, and the coordinates of

the foci are (V<r A

9, 0).

OFThe ratio -----

(thai is, the ratio of the distance of the focus

from the center to the distance of either vertex from the center)

is called the eccentricity of the ellipse and is denoted by t>. lint

M'^VtfCjS, (7)

Vrta

i*and hence c - - -

> (8)\ /

whence it followa that the eccentricity of an ellipse is alwaysless than unity.

Similarly, any equation in form (4), in which I'2 > a2

, representsan ellipse with its center at 0, its major axis on 6>J", and its

minor axis on CLY. Then the vertices are the points (0, &),

~the foci are the points (0, vV

*),and e =

In either case the nearer the foci approach coincidence, the

smaller e becomes and the more nearly b = a. Hence a drde

inat/ le considered an an ellipse with dointxdfnt fovi and equal axes.

Its eceentrieity is, of course, zero.

EXERCISES

Plot tho following ellipses, finding the vertices, the Toci, and the

eceentrieity of eaoh :

1. 9 ^a + 1 if 144. 3. 3 tc8+ 4 ?/ 2.

2. 9 w!8 + 4 ?/ 30. 4. 2 a;

3+ 3 ya s= 1.

6. Find tho equation of the ellipse winch lias its foci at tlie points

( 2, 0) and((>, 0) and which has the sum of the dis lances of any

point on it from the foci equal to 10.

6. Find the equation of the ellipse having its food at the points

(0, 0) and (0, )and having the length of its major axis equal to 7.

32. Hyperbola. The locus of a point the difference of whose

, distances from two faced points is constant is called a hyperbola.

The two fixed points are called the foci*

88 ALGEBRAIC FUNCTIONS

Let F and F' (Fig. 30) be the two foci, and lot tho distance

F'F bo 2 0. Let the straight line determined by F' and F bo

taken as the axis of .P, and the middle point of F'F be taken

as 0, the origin of coordinates, and draw the axis O Y. Thenthe coordinates of F 1 and

F are respectively ( <?, 0)and (c, 0).

Let jP(a;, ?/) be any point

of the hyperbola and 2 a

represent the constant dif-

ference of its distances from

the foci. Then, from the

definition of the hyperbola, . ,

the difference of the dis-'

tances F'P and FP is 2 a,

and from the triangle F'PFii is evident that 2 < 2, for the

difference of any two sides of a triangle IH loss than the third

side ; whence a < a.

1?ia * 80

By 27,

and

whence either

;*+;

2 (1)

or V(* + 0'+^-V(?:::^+7-2a, (2)

according as JRP or ^"/* in the groatcsr diHtance.

Clearing either (1) or (2) of radicals, wo obtain tho aamo

result :

Dividing (3) by a* V, we have

i it _ "1 fA.\' "T a* * \*J

cr r

But since <c,2 *

is a negative quantity whiuh may bo

denoted by fta, and (4) becomes

HYPJEEBOLA 89

Conversely, if the coordinates of any point P satisfy (5), it

can be shown that the difference of the distances F'P and FPis 2 a, and hence P is a point of the hyperbola.

Solving (5) for y in terms of #, we have

y = -^i?-d\ (6)

In this equation we may assume for x only values that are

numerically greater than a, as any other values give imaginaryvalues for y. Hence there are no points of the hyperbola be-

tween the lines t = a and x = a. The hyperbola is symmetricalwith respect to OX.

As the values assigned to x increase numerically, the corre-

sponding points of the hyperbola recede from the axis OX. Wemay, however, write (6) m the form

5?

(7)

Now if yxand i/a

are the ordinates of points of (7) and of

the straight lines y = -x respectively, thenw 7

oa

whence Lim (#a

Hence, by prolonging the straight lines and the curve indefi-

nitely, we can make them come as near together as we please.

Now, when a straight line has such a position with respectto a curve that as the two are indefinitely prolonged the dis-

tance between them approaches zero as a limit, the straight line

is called an asymptote of the curve. It follows that the lines

y = -x and y => as are asymptotes of the hyperbola (Fig. 39).(t CL

If we had solved (5) for a in terms of y, the result would

have been

(8)

90 ALGEBRAIC FUNCTIONS

from which it appears that all values may be assigned to #, and.

that OT is also an axis of symmetry of the hyperbola.

The points A' and A in which one axis of the hyperbola inter-

sects the hyperbola are called the vertices, and the portion of the

axis extending from A' to A is called the transverse axis. The

point midway between the vertices is called the center; that is,

is the center of the hyperbola, and it can readily be shownthat any chord of the hyperbola which passes through O is

bisected by that point. The other axis of the hyperbola, whichis perpendicular to the transverse axis, is called the conju-

gate axis. This axis does not intersect the curve, as is evident

from the figure, but it is useful in fixing the asymptotes andthus determining the shape of the curve for large values of x.

From the definition of 5, c = Va2+ 62,and the coordinates of

the foci are (V 2+52, 0). Therefore

If we define the eccentricity of the hyperbola as the ratio

OF , ,_

> we havea =^+* (10)

a quantity which is evidently always greater than unity.

Similarly, the equation 2

is the equation of a hyperbola, with its center at 0, its trans-

verse axis on OF, and its conjugate axis on OX. Then the ver-

tices are the points (0, 5), the foci are the points (0, V&2+a2

),

the asymptotes are the straight lines y= -#, and e = 7"Cv

If 5 = a, in either (5) or (11), the equation of the hyperbolaassumes the form

a2

-2/2=a2

or y-a^=:aa, (12)

and the hyperbola is called an equilateral "hyperbola. The equa-tions of the asymptotes become y =* x ; and as these lines

are perpendicular to each other, the hyperbola is also called a

rectangular hyperbola.

CURVES

EXERCISES

91

Plot the following hyperbolas, finding the vertices, the foci, the

asymptotes, and the eccentricity of each .

2. 9o;a

4?/a =36. 5.

3. 32/a -2jca =6 6.

7. Find the equation of the hyperbola having its foci at the points

(0, 0) and (4, 0), and the difference of the distances of any point onit from, the foci equal to 2.

8. The foci of a hyperbola are at the points ( 4, 2) and (4, 2),

and the difference of the distances of any point on it from the foci

is 4. Eind the equation of the hyperbola, and plot.

33. Other curves. In the discussion of the parabola, the ellipse,

and the hyperbola, the axes of symmetry and the asymptoteswere of considerable assistance in constructing the curves ; more-

over, the knowledge that there could be no points of the curve

in certain parts of the plane decreased the labor of drawingthe curves. We shall now plot the loci of a few equations,

noting in advance whether the curve is bounded in any direc-

tion or has any axes of symmetry or asymptotes. In this waywe shall be able to anticipate to a con- ^siderable extent the form of the curve.

Ex. 1. (y + 3)a =

(x-

2)a(a; + 1).

Solving for y, we have

In the first place, we see that the only

values that may be assigned to x are greater

than 1, and hence the curve lies entirely on

the positive side of the line x = 1. Further-

more, corresponding to every value of a?, there

are two values of y which determine two points at equal distances from

the line y = 8. Hence we conclude that the line y 8 is an axis of

symmetry of the curve.

Assigning values to x and locating the points determined, we draw the

curve (JFig. 40).

< 49

92 ALGEBKAIO FUNCTIONS

Ex. 2. xy = 4.

4.

Solving foi y, we have y = -x

It is evident, then, that we may assign to x any real value except xero,

in which case we should be asked to divide 4 by 0, a process that cannot

be carried out. Consequently, there can be no point of the curve on the

line # = 0; that is, on OY. We may,however, assume values for x as near

to zero as we wish, and the nearer theyare to zero, the nearer the corresponding

points are to OF; but as the points

come nearer to OY they recede alongthe curve. Hence OF is an asymptote x^of the curve.

If we solve for x, we have

Fm. 41

and, reasoning as above, we conclude

that the line y = (that is, the axis OX) is also an asymptote of the curve.

The curve is drawn in Fig. 41. It is a special case of the curve

xy = k, where k is a ical constant which may be either positive or negative,

and is, in fact, a ioctangular hyperbola leferiod to its asymptotes as axes.

It is customaiy to say that when the denominator of a fraction is SMTO,

the value of the fraction becomes infinite The curve just constructed

shows graphically what is meant by such

an expiession.

Ex. 3. sry + 2 or + y 1 = 0.

Solving for y, we have

y

from which we conclude that the line

a? = 1 is an asymptote of the curve.

Solving for x, we have

Fio. 422+y

fromwhichwe conclude that the line ?/= 2 is also an asymptote of the curve.

We accordingly draw these two asymptotes (Fig. 42) and the curve

through the points determined by assigning values to either x or y and

computing the corresponding values of the other vai table.

The curve is, in fact, a rectangular hyperbola, with the lines x 1

and y SB 2 as its asymptotes.

CUEVES

Ex.4.a-8X

Solving for y, we have

Vx*5 '

2a~a;

whence it is evident that the curve is sym-metrical with respect to OX The lines xand x = 2 a, corresponding to tlie values of a?

which make the numuiatoi and the denomi-

nator of the fraction under the ladical sign

lespectively zero, divide the plane into three

stups; and only values between and 2 a

can be substituted for y, since all other values

make y imaginaiy. It follows that the curve

lies entirely in the strip bounded by the two

lines x and x 2 a.

By the same reasoning that was used in

Exs 2 and 3, it can be shown that the line

x = 2 a is an asymptote of the curve.

The curve, which is called a cuboid, is drawnin Fig. 43.

EXERCISES

Plot the following curves :

1.2/

a =o;8

2.2/

a =aja

(te + 4).

3.2/

a=4(o;-8).

4. ya =a:a

5cc-f 6.

6. yz= a)

(a?

24).

6. /

a =aj8

FIG. 43

7. ya=

8. 7/2=4 a?.

9. xy 5.

10. 3y oj//=

11. xij 2aj +

34. Theorems on limits. In. obtaining more general formulas

for differentiation, the following theorems on limits will be

assumed without formal proof :

1. The limit of the sum of a finite number of variables is equal

to the. sum of the limits of the variables.

2. The limit of the product of a finite number of variables is

egual to the product of the limits of the variables.

94 ALGEBRAIC FUNCTIONS

3. The limit of a constant multiplied by a variable is equal to

the constant multiplied by the limit of the variable.

4. The limit of the quotient of two variables is equal to the

quotient of the limits of the variables, provided the limit of the

divisor is not zero.

35. Theorems on derivatives. In order to extend the processof differentiation to functions other than polynomials, we shall

need the following theorems :

1. The derivative of a constant is zero.

This theorem was proved m 8.

2. The derivative of a constant times a function is equal to the

constant times the derivative of the function.

Let u be a function of x which can be differentiated, let c be

a constant, and place ,. _ .* U -^ G&6i

Give x an increment Ax, and let AM and Ay be the corre-

sponding increments of u and y. Then

Ay = c (u + AM) cu =s c AM.

TT Ay AMHence - =<,

Ax Axand, by theorem 3, 34,

T . Ay T . AMLim -T- C Lim--

Ax Ax

Therefore ~i~ c T~ax dx

by the definition of a derivative.

Ex. 1. y = 5(a;

3 + 3 xz + 1).

3. The derivative of the sum of a finite number of functions is

egual to the sum of the derivatives of the functions.

Let M, v, and w be three functions of x which can be differen-

tiated, and let

DERIVATIVES 95

Give x an increment Aa;, and let the corresponding increments

of u, t>, w, and y be AM, Av, Aw, and A#. Then

Ay = (M H- Aw + v + At) + w + Aw) (w + v + w)

= Aw + Aw + Aw ;

1 Ay AM,

Aw,Aw

whence ^ =-7 + T~~ + 7A# Aa; Aa; Aa;

Now let Aa; approach zero. By theorem 1, 34,

, . Ay T . AM, T A0 . 7 Aw

Lini ^ = Lim - - + Lim - + Lim -;

Aa; Ax Ax Ax

that is, by the definition, of a derivative,

dy __ du dv dw

duo dx dx dx

The proof is evidently applicable to any finite number of

functions.

Ex. 2. y = x* - 3 xs + 2 xz - 7x.

4. The derivative of the product of a finite number offunctions

is equal to the sum of the products obtained "by multiplying the

derivative of each factor by all the other factors.

Let u and v be two functions of x which can be differentiated,

andlety = uv.

Give x an increment Aa;, and let the corresponding increments

of u, v, and y be Aw, Av, and A#.

Then Ay (it +Aw) (v + Av) uv

= u Av + v AM +AM Av

, Ay Av Au . AM Aand = w * + T~ + T-" ^Aa? Ax Ax Ax

96 ALGEBRAIC FUNCTIONS

If, now, Ax approaches zero, we have, by 34,

r Aw T . Aw, T Aw

, T . AM T . ALim -~ = u Lim - + v Lim - + Lim - Lim Aw.Ao; Aa; Aa; Are

But Lim Aw = 0,

-. ,, P dy dv,

duand theretore -^- = w 4- w

ax ax ax

Again, let y = uvw.

Regarding uv as one function and applying the result already

obtained, we have

dy dw, d(uv)-JL = m - + w-+

dx ax at

dw \ du.

<= uv-=--)-w\u + vdx \_ dx ttu?

dw dv du= UV + UW -f VW -r-ax dx dx

The proof is clearly applicable to any finite numbers of factors.

Ex. 3. y = 3a:-

dx

=(3 x

-5) (z" + 1) (3 a;

2) + (3 x

-5)a;

8(2 a;) + (a:

2 + I)*8(3)

= (18 a;8 - 25 a;

2 + 12 a:-

15)a:2

5. The derivative of a fraction is equal to t7ie denominator times

the derivative of the numerator minus the numerator times the deriva-

tive of the denominator, all divided ty the square of the denominator.nt

Let y -> where u and v are two functions of x which can bev

differentiated. Give x an increment Aa;, and let Aw, Av, and A?/

be the corresponding increments of w, v, and y. Then

. _ u +Aw _ u _ v Aw w Aw

v-f-A'y w wa+'uAw

An Az>w---u-and % = A.;

A^^Ao: v

a+ w Aw

DERIVATIVES 97

Now let A* approach zero. By 34,

T . A-M T . A?Jv Liin --- u Lim -

A/ A Aa;Lim - =

,, T . , >

Aa; tr+vLmiAv

du dv01 _________ rilj

_l rty ,

efa/ <fcc difo

whence - =dx v*

T8 1

Ex.4. y = ~ ~Jx* + l

(^_(a,2 + l)(27:)-(a;

2 -l)2r_ 4 .1?

dx (x* + I)3

(a,'J + I)

2

6. TJie derivative of the nth power of a function is obtained ly

multiplying n times the (nT)th power of the function ly the

derivative of the function.

Let y = un, where u is any function of x which can be differ-

entiated and n is a constant. We need to distinguish four cases :

CASE I. When n is a positive integer.

Give x an increment Aa, and let AM and A?/ be the corre-

sponding increments of u and y. Then

Ay = (u 4-Aw)" un ;

whence, by the binomial theorem,

A?/ n_iAw ,n(n 1) B _ 2A AM

,

^L = nun l ~ H ft

y ww 2Au- h .

Aa; Aa? 2 Aa;'

Aa;

Now let Aa;, AM, A?/ approach zero, and apply theorems 1 and

2, 34. The limit of ^ is ^ the limit of ~ is ^*, and theAa; a* Aa; ajc

limit of all terms except the first on the right-hand side of

the last equation is zero, since each contains the factor AM.

Therefore d du^. = nwn

~ 1 -T- <

ax dx

08 ALGEBKAIC FUNCTIONS

CASE II. When n is a positive rational fraction.

ny

Let n = where p and q are positive integers, and place

By raising both sides of this equation to the gth power, we have

y=u.Here we have two functions of x which are equal for all

values o x.

Taking the derivative of both sides of the last equation, we

have, by Case I, since p and q are positive integers,

^-1^==*-^.yy dx P dx

Substituting the value of y and dividing, we have

f^itf-1

**.* dx q dx

Hence, in this case also,

dy _ ,du-^- nun~ l

dx ax

CASE III. When n is a negative rational number.

Let n m, where m is a positive number, and place

ys=U~m s= --9 um

Then ^= **(by 5)dx uzm

mum~ l

=(by Cases I and II)u

mu~ m~^~rdx

Hence, ui this case also,

d_ _i^wdx dx

DERIVATIVES 99

CASE IV. When n is an irrational number.

The formula is true in this case also, but the proof will not

be given.

It appears that the theorem is true for all real values of n.

It may be restated as a working-rule in the following words:

To differentiate a power of any quantity, bring down the exponentas a coefficient, write the quantity with an exponent one less, and

multiply ly the derivative of the quantity.

Ex. 5. y = (xs + 4 x* - 5 x + 7)

s.

^ = 3(a:

8 + 4 a2 - 5 x + 7)2

-f (a;3 + 4 xz - 5 x + 7)

tlJG U3S

= 3 (So;2 + Bx - 5) (.e

3 + 4a:2 - 5 x + 7)3.

Ex. 6. y = Vtf + -. = ^ + x-*

dx 3

2 3*'

Ex. 7. y = (a: + l)Va;8

=(z + 1) [ J (a;

2 + I)" * - 2ar] + (a

2 +

1

(a2 + !)

2 a8 + x 4- 1

Y /Li

100 ALGEBRAIC FUNCTIONS

7. If y is a function of x, then x is a function of y, and the

derivative of x with respect to y is the reciprocal of the derivative

of y with respect to x.

Let A# and A?/ be corresponding increments of x and y. It is

immaterial whether b*.x is assumed and A?/ determined, or A#is assumed and Aa; determined. In either case

A.r 1

A^~"Ay'Aa;

. _. A,r 1whence Lun =

A ;

Ay T A?/' Lim -Aas

dx 1that is,

-- = -

8. If y is a function of u and u v' function of a*, then y is

a function of x, and the derivative of y with respect to x is equal

to the product of the derivative of y with respect to u and the

derivative of u with respect to x.

An increment Aa; determines an increment Aw, and this in turn

determines an increment Ay, Then, evidently,

A?/ __ A?/ AwAx Aw A:#

, T . Aw , . A?/ T . AMwhence Lim ~ = Lim - Lim . ;

A:B A?t Arc

, , , . dy dy duthat is, ^ .X. .

dx du dx

Ex. 9. w a u3 + 8 u + 1, whoro w = -

1+

The same result is obtained by fmbHtituting in the expression for y the

value of u in terms of a; and then differentiating.

DERIVATIVES 101

36. Formulas. We may now collect our formulas of differen-

tiation in the following table :

^ = 0, (1)dx

rfco = tf *, C2)dx dv

d(u4-v) _du dv n

dx dx dx

d Cuv) dv,

du , ,,

\ - UT + V 1T' (4)dx dx dx

, /?/\ du dvd(-) v--- M\v/ dx dx

dy^y. d*t /g-)

dx du dx ^ ^

dy

^ = (9)dx dx

du

Formula (9) is a combination of (7) and (8).The first six formulas may be changed to corresponding for-

mulas for differentials by multiplying both sides of each equation

by dx. They are 7 n _<- vJ J da = 0, (10)

(11)

Zw, (12)

d (uv) udv + v du, (1 3)

w\ vdu udv

102 ALGEBRAIC FUNCTIONS

EXERCISES

Find -~ in each of the following oases :

2. y s (a;

2 - 2 1 + 3) (a:3 +60: + 9).

14.H-l"**

7 .

8 X?/ = (4 jrfl + 3 x + I)

3-

4 X* + 1.

16.> re V<.)

17 ''/= (' +

18.

37. BijEferentiatlon of implicit functions. Consider any equa-tion containing two variables a? and

,?/,If one of them, EH jr, in

chosen as the independent variable and a value is assigned in

it, the values of y are determined, Hence tho given equationdefines y as a function of rr. If the equation is solved for y in

terms of a, y is called an earpKoit function of x, If the equationis not solved for y, y is (jailed an implidt function of #* For

example, ^+8aM-4y + 4^ + 2^ + 4^0,which may be written

#9+ (4 aj + 2)y + (8 .-K

9-|- 4 4- 4) 0,

defines y as an implicit function of a?.

If the equation is solved for y, the renult

# 2j lVa^T8expresses y as an explicit function of #.

IMPLICIT FUNCTIONS 103

If it is required to find the derivative of an implicit function,

the equation may be differentiated as given, the result being an

equation which may be solved algebraically for the derivative.

This method is illustrated in the following examples :

Ex. 1. z2 + 2 = 5.

If x is the independent variable,

that is, 2 x + 2 y- = 0,dx

whence Jl = -.dx y

Or the derivative may be found by taking the differential of both sides,

as follows: + /) = d(5) = 0;

that is, 2 xdx + 2 ydy = 0,

whence ^=_.dx y

It is also possible first to solve the given equation for y, thus .

y = V5-ff2;

whencerfa; V5 a:

8

a result evidently equivalent to the result previously found.

The method of finding the second derivative of an implicitfunction is illustrated in the following example :

Ex. 2. Find ^f if aa + f = 5(IJS

We know from Ex. 1 that ^L =_ 2 .

dx y

Therefore fSUfLftdor ax \y/

.*-*(!)

_ ?ya + a-

8 _ 5

f 2/8

'

since ^2 + a;

2 =5, from the given equation

104 ALGEBRAIC FUNCTIONS

EXERCISES

"Find ~ from each of tho following equations :

CtJC

1. a* + / - 3 *y = 0. 3. v/= ~~-j -|- //

2. aj2// -f- 4 V = 8 a9

. 4. V // + ' + V^ ,r -- .

Pind ^ andl

-~. from each oJ' tho following (wuiationa :

6. 2xa

6. 4 a-3

9v/a =3C.

7 . ai+jf-a.io -

8. a* + * = <A 11.

38. Tangent line. Let J^C^'j, ^O bo a clioHou point, of any

curve, and lot (^) bo the value of '/- when aj=sw(

. and ;/-://,.V ^V v / // >* ^ ^

/x..\ VWi <M

Then f-^J is the slopo oC tho curve at, iho point J^ and also\dx/\

the slopo of tho tangout line ( 1H) to tho ourvo at that point,

Accordingly, tho equation of the tang-out line at J{ m ( 15)

Ex. 1. Find tho equation (if tho lungojit lino to tho parabola /8 ss JJ .r at

tho point (3, ).

By differentiation wo have

whence //SB.'-.,. ti //

Hence, at tho point (3, JJ),tho Hlopn of tho tangont lino is J, and itH

equation is y-3-1 (*-)or ar 2

?/ + 3 = 0.

The angle of mteraeotion of two curves is tho angle betwoon

their respective tangontu at the point of iutenwotiun. Themethod of finding tho anglo of intersection i illustrated in the

example on. tho following pago*

TANGENT LINE 105

Ex. 2. Find the angle of intersection of the cncle #2 + #2 = 8 and of

the parabola ar2 = 2 y. v

The points of intersection are Pl (2, 2)

and />(- 2, 2) (Fig 44), and fioiu the sym- ^nietry of the diagram it is evident, that the

angles of intersection at Pland Pz aie the

same.

Diffeientiating the equation of the circle,

we have 2 JT + 2 y = 0, whence =;'

(Ijc ux yand differentiating the equation of tlTb pa-

labola, we find = i. Fid 44

Hence at Pithe slope of the tangent to the circle1 is 1, and the slope

of the tangent to the parabola is 2

Accordingly, if /? denotes the angle of intersection, by Ex 11, p 35,

ory

= tan~ 1 3.

EXERCISES

1. Find the equation of the tangent lino to the ourve x8

+ 16 y - 8 = at the point (2, 2).

2. Find the equation of the tangent lino to the p,urve 5 a;2 4 a-

2//

= 4 y8at the point (2,1).

3. Find the point at which the tangent to the curve 8 y = a;8 at

(1, &) intersects the curve again.

4. Find the angle of intersection of the tangents to the curve

yz = x9 at the points for which x = 1

9T If

5. Show that the equation of the tangent to the ellipse ~j + ^ = 1

at the point (a^, y,)is + -M = L

6. Show that the equation of the tangent to the hyperbola

|-

|j= 1 at the point fo, ^) is^ - %& = I.

Y. Show that the equation of the tangent to the parabola ?/ = kx1:

at the point (xlt y^ is y$ = ^ (as + a 1

,).

Draw eacli pair of the following curves in one diagram and deter-

mine the angles at which, they intersect :

106 ALCKEBRAK! FUNCTIONS

8. a:9 + y

j = S, a-9 + if

- U ^ H- 4>/- f> 0.

9. a;a = 3 #, U //

a 8 a-.

10. yfl ~ 4

';<* 'I' //"

-- fi

11. y = 2.r, .<//- 18.

12. x 4//

J 0, .r - --1 j' 4 if 0.

13. aja +y9 ~2,l

>,.<'a-//- ".

39. The differentials <?#, dy, ds. On any given <uw li-t tin-

distance from somo Jixod initiul point moasmvd alnnjjf the cur\i

to any point P bo donoiwd by , whcro is posit.ivt* if /' lifs in

one direction from the initial point, and negative if /' lies in I In-

apposite direction. Tho (jlit)ico of tho ^positive direction is purely arbitrary. ./

We shall take as the positive direc-

tion of the tangent that which shown

the positive direction of the ourvo,

and shall denote the angle between //the positive direction of (L\ and the

'

positive direction of the tangent by 0,

Now for a fixed curve and a fixed

initial point the position of a point /'p, (Ji

is determined if a is given. Hence &

and y, the coordinates of J\ are functions of which in

are continuous and may bo differentiated. \V Hhall now Mhuw tlmt

dx , dy . ,

-7- COS G>,- aa BUI O,

ds </

Let arcP0=Afl (Fig. 45), whenj 7* and Q are so <hnm thai

As is positive. Then 1'R AJJ and A*<y A^/, and

'/f

As arc /'( tiro /%; chord /</

" 00

Aa

chord PQ

MOTION IN A CURVE 107

We shall assume without proof that the ratio of a small chord

to its arc is very nearly equal to unity, and that the limit of

"=1 as the point Q approaches the point P along the

curve. At the same time the limit of RPQ = <f>. Hence, taking

limits, we have dx d

^ = cos & ^ = sm ^ (1)ttrO l*O

If the notation of differentials is used, equations (1) become

dx = ds cos 0, dyds' sin $ ;

whence, by squaring and adding, we obtain the important equation

ds = dx + dy . (2)

This relation between the differentials of #, y, and s is often rep-

resented by the triangle of Fig. 46. This figure is convenient as a

device for memorizing formulas (1) and (2), but it should be borne

m mind that RQ is not rigorously yequal to dy ( 20), nor is PQ rigor-

ously equal to ds. In fact, RQ = Ay,and PQ = As

; but if this triangle is

regarded as a plane right triangle,

we recall immediately the values of

sin^>, cos<, and tan< which have

been previously proved.40. Motion ia a curve. When a o JL

, , . .IT FIG. 46body moves m a curve, the discus-

sion of velocity and acceleration becomes somewhat complicated,

as the directions as well as the magnitudes of these quantities

need to be considered. We shall not discuss acceleration, but

shall notice that the definition for the magnitude of the velocity,

or the speed, is the same as before (namely,

dsV=S 3I'

dt

where s is distance measured on the curved path) and that the

direction of the velocity is that of the tangent to the curve.

108 ALGEBRAIC IUNCTIONS

Moreover, as the body moves along a curved path through a

distance PQ=&s (Fig. 47), x changes by an amount PII&K,and y changes by an amount

RQ=&y. We have then

Lim = -^ = <y = velocity ofA at

the body in its path,

Ax dx ,

Lim =-7-= vx = component

At dt

of velocity parallel to OX,

Lim^ =^ = vv = componentAt dt

J rFIG 47

of velocity parallel to Y.

Otherwise expressed, v represents the velocity of P, vr the

velocity of the projection of P upon 0-3T, and vvthe velocity

of the projection of P on OY.

Now, by (8), 36, and by 39,

_ dx _ dx dsx

dt ds dt

= V COS <, (1), dy dy ds

and v. .

~ = -*y

dt ds dt

= v sm <p. (2)

Squaring and adding, we have

v*=v*+ V;. (3)

Formulas (1), (2), and (3) are of especial value when a par-ticle moves in the plane XO F, and the coordinates x and y of

its position at any time t are each given as a function of t. Thepath of the moving particle may then be determined as follows :

Assign any value to t and locate the point corresponding to

the values of x and y thus determined. This will evidently bethe position of the moving particle at that instant of time. Inthis way, by assigning successive values to t we can locateother points through which the particle is moving at the corre-

sponding instants of time. The locus of the points thus deter-mined is a curve which, is evidently the path of the particle.

MOTION IN A CURVE 109

The two equations accordingly represent the curve and are

called its parametric representation, the variable t being1

called a

parameter.* By (9), 36, the slope of the curve is given bythe formula jal

dx d vx

dt

In. case t can be eliminated from the two given equations, the

result is the (#, /) equation of the curve, sometimes called the

Cartesian equation; but such ehmmation is not essential, and

often is not desirable, particularly if the velocity of the particlein its path is to be determined.

Ex. 1. A particle moves m the plane XOY so that at any time t,

a; = a + bt, y = c + dt,

where a, &,' c, and d are any real constants Determine its path and its

velocity in its pathTo determine the path we eliminate t from the given equations, with

the resultd f ..

y c = -(x d),

the equation of a straight line passing through the point (a, c) with the

id

slope-

In this case the path may also be determined as follows Fiom the given

equations we find dx l dt, and dy ddt\ whence = '- As the slope of

/ rA*'

the path is always the same (that is, -I, the path must be a straight line

which passes through the point (a, c) the point determined when t = 0.

To determine the velocity of the particle in its path we find, by differ-

entiating the given equations,

dx , di/ .

v*=jr l> v =dt= d >

whence, by (3), v = V&2H- d\

Hence the particle moves along the straight line with a constant velocity.

* It may be noted m passing that the parameter in the parametiic represen-tation of a curve is not necessarily time, but may be any third variable in termsof which a and y can be expressed.

110 ALGEBRAIC FUNCTIONS

Ex. 2. If a projectile starts with an initial velocity v in an initial direc-

tion which makes an angle a with the axis of x taken as horizontal, its

position at any time t is given by the parametnc equations

x v t cos a, y = v t sin a \ gtz.

JTind its velocity in its path.dx

We have vx = = v cos a,at

dyvv=~ft

~'

Hence = Vi> 2 gv t sin a + g*P.

EXERCISES

1. The coordinates of the position of a moving particle at anytime t are given by the equations x = 2

1, y ts Determine the path.

of the particle and its speed in its path

2. The coordinates of the position of a moving particle at anytime t are given by the equations x t

2, y = t + 1 Determine the

path of the particle and its speed in its path.

3. The coordinates of the position of a moving particle at anytime t are given by the equations x = 2 1, y == f t $ t*. Determinethe path of the particle and its speed in its path.

4. At what point of its path will the particle of Ex. 3 be movingmost slowly ?

5. The coordinates of the position of a moving particle at anytime t are given by the equations x = ? 3, y t

9 + 2. Determinethe path of the particle and its speed in its path

6. The coordinates of the position of a moving particle at anytime t are given by the equations x = 4 &, y 4 (1 t)

2. Determine

the path of the particle and its speed in its path.7. Find the highest point in the path of a projectile.8. Find the point in its path at Trhich the speed of a projectile

is a minimum.

9. Find the range (that is, the distance to the point at whichthe projectile will fall on OX), the velocity at that point, and theangle at which the projectile will meet OX.

10. Show that in general the same range may be produced bytwo different values of a, and find the value of which producesthe greatest range.

11. Find the (a, y) equation of the path of a projectile, and plot.

VELOCITIES AND RATES 111

41. Related velocities and rates. Another problem of some-what different type arises when we know the velocity of onepoint iii its path, which may be straight or curved, and wish tofind the velocity of another point which is in some way con-nected vviUi tho first but, in general, describes a different path.The method, in general, is to form an equation connecting thedistances traveled by the two points and then to differentiate

tho equation thus formed with respect to the time t. The resultis an equation connecting the velocities of the two points.

Ex. 1 A lump is 00 ft. above tho ground. A stone is let drop from a

point on tho same lovol us the lamp and 20 ft, away from. it. Find the speedof the stone's shadow on the groundat the end of 1 HOC., assuming that the

distance traversed by a falling body in

the time t is 1(5 <a.

Lot A C (Fig, 48) be tho surface of

tho ground which is assumed to be a

homontal plane, L the position of the

lamp, the point from which the stone

was dropped, and S the position of the _stone at any time t. Then Q is the posi- ^I0 4gtion of the shadow of S on the ground,

LSQ being a straight line. Let OS = x and BQ,= ?/.Then L = 20, BO= 60,

and BS = 60 a?. In the similar triangles LOS and SBQ,

whence y SB~ 20. (2)7

^3

We know x =16 <a, whence~ = 82 1

;andwish to find?, the velocity of Q.

at at

Difterontiating (2) with respect to t, we have

dt x* dt

When / s= 1 sec,, x =s 16, and ~ =s 82; whence, by substitution, we find

at

~& ss 150 ft. per second.dt

The result is negative because y is decreasing as time goes on.

In 6 and 11, if the rate of one of two related quantitieswas known, we were able to find the rate of the other quantity.

112 ALGEBRAIC FUNCTIONS

This type of problem may also be solved by the same method

by which the problem of related velocities has been solved.

We shall illustrate by taking the same problem that was used

in 11.

Ex. 2. Water is being poured at the rate of 100 cu. in. per second into

a vessel in the shape of a right circulai cone of radius 3 in. and altitude

9 iu. Required the rate at which the depth of the water is increasing when

the depth is 6 in.

As in 11, we have V = ^ irha

;

dV1

, zd1i

whence -^= * ^

We have given= 100, h = 6

;from which we compute

^ = 25 =796 .

dt v

EXERCISES

1. A point is moving on the curve y2= sc

s The velocity alongOX is 2 ft. per second What is the velocity along OY when x = 2 ?

2. A ball is swung in a circle at the end of a cord 3 ft long so

as to make 40 revolutions per minute. If the cord breaks, allowingthe ball to fly off at a tangent, at what rate will it be receding fromthe center of its previous path 2 sec after the cord breaks, if noallowance is made for the action of any new force ?

3. The mside of a vessel is in the form of an inverted regular

quadrangular pyramid, 4 ft square at the top and 2 ft deep. Thevessel is originally filled with water which leaks out at the bottomat the rate of 10 cu. in. per minute. How fast is the level of the

water falling when the water is 10 in. deep ?

4. The top of a ladder 20 ft long slides down the side of a ver-

tical wall at a speed of 3 ft per second. The foot of the ladder slides

on horizontal land. Find the path described by the middle point of

the ladder, and its speed in its path.

5. A boat with the anchor fast on the bottom at a depth of 40 ft.

is drifting at the rate of 3 mi per hour, the cable attached to theanchor slipping over the end of the boat At what rate is the cable

leaving the boat when 50 ft of cable are out, assuming it forms a

straight line from the boat to the anchor ?

GENERAL EXERCISES 113

6. A solution is being pouied into a conical filter at the rate of

5 cc per second and is running out at the rate of 2 cc. per second.

The radius of the top of the filter is 8 cm and the depth of the filter

is 20 cm. Find the rate at which the level of the solution is risingin the filter when it is one third of the way to the top.

7 . A trough is in the form of a right prism with its ends isosceles

triangles placed vertically. It is 5 ft long, 1 ft across the top, and8 in. deep It contains water which leaks out at the rate of 1 qt.

(57| cu. in ) per minute Find the rate at which, the level of the

water is sinking in the trough when the depth is 3 in.

8. The angle between the straight lines AB and BC is 60, andAB is 40 ft. long. A particle at A begins to move along AB towardB at the rate of 5 ft per second, and at the same time a particle at

B begins to move along BC toward C at the rate of 4 ft per second. Atwhat rate are the two particles approaching each, other at the endof 1 sec *

9. The foot of a ladder 50ft. long rests on horizontal ground, andthe top of the ladder rests against the side of a pyramid which makesan angle of 120 with the ground. If the foot of the ladder is drawn

directly away from the base of the pyramid at the uniform rate

of 2 ft. per second, how fast will the top of the ladder slide downthe side of the pyramid ?

GENERAL EXERCISESPlot the curves :

1. 3 a2+ 7 f = 21 9.?/2

(4 + or2

)= cc

2

(4- a2

).

2. 4?/a == 9.r. z

a, x*10. ?/- = a;

a- .

3. 9a;2 y*=lG. a + x

4.2/

a- 2 y = a;8 + 2 a? - 1.

n

_ 8 a8 12.y -rf+4a"' IS.

6.

7, (7/-aJ)2=9-a!

2 > ==

^T4*8. (* + yf f(y + 2). 15. xY + 36 = 16f

Find the turning-points of the following curves and plot the curves

16. y = (2 + oO(4-

a-,)

2. = (x

-I)

3

17. y = (x + 3)2

(a:-

2).

' JJB + 1

'

a 03,- at

x

114 ALGKEBBAIC FUNCTIONS

20. Find the equation of the tangent to the curve f = xz~-

/ 3 a 6a\the point (

-g-> ~6/

21. Find the equation of the tangent to the curve x* + y* = *

at the point (a;.,^, 3^).

22. Prove that if a tangent to a parabola ?/2 = 7o has the slope ?;?,

its point of contact is (j-5, ^-) and therefore its equation is

A 4 VTE 7/t' J& tfvj

a 2JE ?/

23. Prove that if a tangent to an ellipse + ~ = 1 has the slope m,ft U

its point of contact is(

* m,^ ,

&

2 )and therefore

^\ Vrt2

?^2 + 62 Varm2 + 67

its equation is y = wcc Va2m2 + i2

24. Show that a tangent to a parabola makes equal angles with

the axis and a line from the focus to the point of contact.

25. Show that a tangent to an ellipse makes equal angles with the

two lines drawn to the foci from the point of contact

Find the angles of intersection of the following pairs of curves .

26. ,= ~

28. a;2=

29. a:2-4y-4=0, a? + 12y - 36 = 0.

30. 2/! =o;8

} 2/!

=(2-a;)8.

31.

32.

33. The coordinates of a moving particle are given by the equa-

tions x tf

3, y = (1 tf

2

)^.Find its path and its velocity in its

path.

34. A particle moves so that its coordinates at the time t are

2x = 2t, y = . . Find its path and its velocity in its path.

35. A projectile so moves that x = at, y = bt ^g^. Find its

path and its velocity in its path.

GENERAL EXERCISES 115

36. A body so moves that x = 2 + tf$, y = 1 -f t. Find its pathand its velocity in its path.

37. A particle is moving along the curve if 4. x; and when as = 4,

its ordmate is increasing at the rate of 10 ft. per second. At what

rate is the abscissa then changing, and how fast is the particle movingin the curve ? Where will the abscissa be changing ten times as fast

as the ordmate ?

38. A particle describes the circle o;a+ya =a2 with a constant

speed v . Find the components of its velocity.

39. A particle describes the parabola y2 = 4 ax in such a way that

its a-component of velocity is equal to ct Find its y-component of

velocity and its velocity in its path.

40. A particle moves so that x = 2 1, y = 2 "Vt & Show that it

moves around a semicircle in the time from t = to t = 1, and find

its velocity in its path during that time.

41. At 12 o'clock a vessel is sailing due north at the uniform rate

of 20 mi. an hour. Another vessel, 40 mi. north of the first, is sailing

at the uniform rate of 15 mi an hour on a course 30 north of east.

At what rate is the distance between the two vessels diminishing at

the end of one hour ? What is the shortest distance between the

two vessels ?

42. The top of a ladder 32 ft. long rests against a vertical wall,

and the foot is drawn along a horizontal plane at the rate of 4 ft.

per second in a straight line from the wall. Find the path of a

point on the ladder one third of the distance"from the foot of the

ladder, and its velocity in its path.

43. A man standing on a wharf 20 ft. above the water pulls in a

rope, attached to a boat, at the uniform rate of 3 ft. per second Find

the velocity with which the boat approaches the wharf

44. The volume and the radius of a cylindrical boiler are expand-

ing at the rate of .8 cu. ft. and .002 ft. per minute respectively. Howfast is the length of the boiler changing when the boiler contains

40 cu. ft. and has a radius of 2 ft. ?

45. The inside of a cistern is in the form of a frustum of a regular

quadrangular pyramid. The bottom is 40 ft. square, the top is 60 ft.

square, and the depth is 10 ft, If the water leaks out at the bottom

at the rate of 5 cu. ft. per minute, how fast is the level of the water

falling when the water is 5 ft, deep in the cistern ?

116 ALGEBRAIC FUNCTIONS

46. The inside of a cistern is in the form of a frustum of a right

circular cone of vertical angle 90. The cistern is smallest, at the

base, which is 4 ft in diameter. Water is being poured in at the rate

of 5 cu ft. per minute. How fast is the water rising in tho cistern

when it is 2 ft deep ?

47. The inside of a bowl is in the form of a hemispherical sur-

face of radius 10 in If watei is running out of it at the iat,o of

2 eu in. per minute, how fast is the depth of the water decreasingwhen the water is 3 in deep ?

48. How fast is the surface of the bowl in Ex 47 being exposed ?

49. The inside of a bowl 4 in deep and 8 in. across the top is in

the form of a surface of revolution formed by revolving a parabolic

segment about its axis Water is running into the bowl at the rate

of 1 cu in per second How fast is the water rising in the bowlwhen it is 2 in deep

?

50. It is required to fence off a rectangular piece of ground to con-

tain 200 sq ft,one side to be bounded by a wall already constructed.

Find the dimensions which will require the least amount of fencing.

51. The hypotenuse of a right triangle is given. Find the othersides if the area is a maximum

52. The stiffness of a rectangular beam varies as the product of thebreadth and the cube of the depth. Find the dimensions of the stiffest

beam which can be cut from a circular cylindrical log of diameter 18 in.

53. A rectangular _plot of land to contain 384 sq. ft. is to be in-closed by a fence, and is to be divided into two equal lots by a fenceparallel to one of the sides What must be the dimensions of tho

rectangle that the least amount of fencing may be required ?

54. An open tank with a square base and vertical sides is to havoa capacity of 500 cu ft Fiud the dimensions so that the cost oflining it may be a minimum

55. A rectangular box with a square base and open at the top isto be made out of a given amount of material. If no allowance ismade for thickness of material or for waste in

construction, what arethe dimensions of the largest box which can be made ?

^

56. A metal vessel, open at the top, is to be cast in the form of aright circular cylinder. If rt is to hold 27 TT cu in

,and the thickness

of the side and that of the bottom are each to be 1 in, what will be theinside dimensions when the least amount of material is used ?

GENERAL EXERCISES 117

57 . A gallon oil can (231 cu in)

is io be made in the form, of a

right chcular cylinder. The material used for the top and the bottom

costs twice as much per square inch as the material used for the

side What is the radius of the most economical can that can be

made if no allowance is made for thickness of material or waste in

construction ?

58. A tent is to be constructed in the form of a regular quadran-

gular pyiauucl Find the ratio of its height to a side of its babe whenthe air space inside the tent is as great as possible for a given wall

surface

59. It is required to construct from two equal circular plates of

radius a a buoy composed of two equal cones having a common base.

Find the radius of the base when the volume is the greatest

60. Two towns, A and I>, are situated respectively 12 mi. and

18 mi. back from a straight river from which they are to get thoir

water supply by means of the same pumping-station. At what pointon the bank of the river should the station be placed so that the least

amount of piping may be required, if the nearest points 011 the river

from A and B respectively are 20 mi. apart and if the piping goes

directly from the pumping-station to each of the towns 9

61. A man 011 one side of a river, the banks of which are assumed

to be parallel straight lines mi apart, wishes to reach a point on

the opposite side of the river and 5 mi. further along the bank If

he can row 3 mi. an hour and travel on land 5 mi. an hour, find the

route he should take to make the trip in the least time.

62. A power house stands upon one side of a river of width I miles,

and a manufacturing plant stands upon the opposite side, a miles

downstream. Find the most economical way to construct the con-

necting cable if it costs m, dollars per mile on land and n dollars a

mile through water, assuming the banks of the river to be parallel

straight lines.

63. A vessel A is sailing due east at the uniform rate of 8 mi.

per hour when she sights another vessel B directly ahead and 20 mi.

away. B is sailing in a straight course S. 30 W at the uniform rate

of 6 mi per hour. When will the two vessels be nearest to each other?

64. The number of tons of coal consumed per hour by a certain

ship is 0.2 + 0.001 v*}where v is the speed in miles per hour. Find

an expression for the amount of coal consumed on a voyage of

1000 mi. and the most economical speed at which to make the voyage.

118 ALGEBRAIC FUNCTIONS

65. The fuel consumed by a certain steamship in an hour is pro-

portional to the cube of the velocity which would be given to the

steamship in still water. If it is required to steam a certain distance

against a current flowing a miles an hour, find the most economical

speed. a 2

66. An isosceles triangle is inscribed in the ellipse ^ 4-~~ =

1,Ct ()

(a > 6), with its vertex in the upper end of the minor axis o the

ellipse and its base parallel to the major axis Determine the lengthof the base and the altitude of the triangle of greatest area whiuli

can be so inscribed.

CHAPTER V

TRIGONOMETRIC FUNCTIONS

42. Circular measure. The circular measure of an angle is the

quotient of the length of an arc of a circle, with its center at

the vertex of the angle and included between its sides, divided

by the radius of the arc. Thus, if 6 is the angle, a the lengthof the arc, and r the radius, we have

-~ CD

The unit of angle in this measurement is the radian, which

is the angle for which a r in (1), and any angle may be said

to contain a certain number of radians. But the quotient - inr

formula (1) is an abstract number, and it is also customary to

speak of the angle 6 as having the magnitude - without using

the word radian. Thus, we speak of the angle 1, the angle -|,

the angle > etc.

In all work involving calculus, and in most theoretical workof any kind, all angles which occur are understood to be ex-

pressed in radians. In fact, many of the calculus formulas wouldbe false unless the angles involved were so expressed. Thestudent should carefully note this fact, although the reason for

it is not yet apparent.From this point of view such a trigonometric equation as

y = sin x (2)

may be considered as defining a functional relation between two

quantities exactly as does the simpler equation y a;3. For

we may, in (2), assign any arbitrary value to x and determine

the corresponding value of y. This may be done by a direct

119

120 TRIGONOMETRIC FUNCTIONS

computation (as will be shown in Chapter VII), or it may be

done by means of a table of trigonometric functions, in which

case we must interpret the value of x as denoting so many radians.

One of the reasons for expressing an angle in circular measure

is that it makes true the formula

*T bill fv *4 xrtxLim - = 1, (3)A-+0 fl

where the left-hand member of the equation is to be read

"the limit of

^-yas h approaches zero as JB

a limit."

To prove this theorem we proceed as

follows:->. Ti '

Let h be the angle A OB (Fig. 49), r the^%\x j //

radius of the arc AB described from as""x^j/

a center, a the length of AB, p the lengthB '

of the perpendicular BC from B to OA,

and t the length of the tangent drawn from B to meet OA

produced in D.

Revolve the figure on OA as an axis until B takes the position

B'. Then the chord BGB'=2p, the arc BAB'=2 a, and the

tangent BfD=tliG tangent BD. Evidently

BD +DB' > BAB' >BCB';

whence t>a>p.

Dividing through by r, we have

r r r

that is, tan h > h > sin h.

Dividing by sin A, we have

cos h sin A'

or, by inverting, cos h < ^ ^ < 1.h

GRAPHS 121

Now as h approaches zero, cos h approaches 1. Hence rft

which lies between cos h and 1, must also approach 1 ; that is,

A - o n

This result may be used to find the limit ofk

as It

approaches zero as a limit For we havel

c\ a fl n /I 'I/

2 sin3 - sin

2 - sm-1 cos h _ 2 2 _ h 2

A~

h I 2 7t

2 2,

. hsm-Now as h approaches zero as a limit, approaches unity,

by (3). Therefore h

r 1 cos h A .

,.

Lira =0. (4)A-O 7i

V '

43. Graphs of trigonometric functions. We may plot a trigo-

nometric function by assigning values to x and computing, or

taking from a table, the corresponding values of y. In so doing,

any angle which may occur should be expressed in circular

measure, as explained m the previous section. In this connec-

tion it is to be remembered that TT is simply the number 3.1416,and that the angle w means an angle with that number of radians

and is therefore the angle whose degree measure is 180.The manner of plotting can be best explained by examples.

Ex. 1. y = a sin fa-

it is convenient first to fix the values of x which make y equal to aero

Now the sine is zero when the angle is 0, IT, 2 w, 3 IT, TT, 2 TT, on, iu

general, kv, where k is any positive or negative integer. To make y = 0,

therefore, we have to place bx = kv, whence

2w IT n TT 2ir 8w* > r- > U, ~) --

, ~j > '.

b b b b b

The sine takes its maximum value + 1 when the angle has the valuesif 5 TT OTT TT 5 TT TT

e>'~2"' ~n~' eifcc< a* *s>*n *kis case> "when x = > , ~, etc. For these

values of or, w = a

122 TRIGONOMETRIC FUNCTIONSO rr .

The sine takes its minimum value 1 when the angle is '-, - . etc.,

q 17_ ii 2

that is, in this case, when x = ' - etc Foi these values of r, y = a.20 2o

These values of x for which the sine is 1 lie halfway between the

values of x for which the sine is

7

FIG 50

These points on the graph are enough to determine its general shapeOther values of x may be used to fix the shape more exactly The graphis shown in Fig 50, with a = 3 and I = 2 The curve may be said to repre-

sent a wave. The distance from peak to peak, -, is the wave length, andthe height a above OX is the amplitude

Ex 2. y = a cos Ix.

As in Ex 1, we fix first the points for which y = Now the cosine of

an angle is zero when the angle is -> , , etc;

that is, any odd

multiple of \> We have, therefore, y = whenAt

26'

STT

26'

Y2b

FIG. 61

Halfway between these points the cosine has its maximum value + 1or its minimum value - 1 alternately, and y = a. The graph is shownin Fig 51, with a = 3and& = 2

GKRAPHS 128

Ex. 3. y = a sin (bx + c).

We have y = when bx + c = 0, TT, 2 TT, 3 rr, c>tc.; that is, when

7 T2ir

]?i&. 52

Halfway between these values of x tho HHIQ has its maximum value + 1

and its minimum value 1 alternately, and y db a. Tho tmvvo is tho

same as in Ex. 1, but is shifted ~ units to the loft (Fig. 52).

Ex, 4. y sin a; + 4 sin 2 ,x.

The graph is found by adding the ordinatcs of tho two curves y = sin x

and y ^ sin 2 x, as shown in Fig. 58.

Y

sin x+i sin 8*0

3?io. 68

EXERCISES

Plot the graphs of the following equations :

1. y us 2 sin 3 a:. 6. y

2. 7/ = 3oos 7 - y8, y

3. Ssinfaj ~V 9.2/

^x 10. y

-j. 11. y

tan 2 SB,

SOOCB.

4. ^2008vers jr.

6. y *a 3 sin(as 2), 12, gin 4. s in

124 TKIGONOMETKIC FUNCTIONS

44. Differentiation of trigonometric functions. The formulas

for the differentiation of trigonometric functions are as follows,

where u represents any function of x which can be differentiated :

d . dusinM = cosM -

dx dx

d ducosM = sinw >

dx dx

d . 2 dutan u = sec u ->

dx dx

A ctn W = -csc2

w^, (4)dx dx

d , du XCNsec u = sec u tan u ( o)

dx dx

d , du X /JNesc u = esc u ctn u (6)

dx dx

These formulas are proved as follows:

1. Let y sin w, where u is any function of a; which may be

differentiated. Give x an increment A and let AM and Ay be

the corresponding increments of u and y. Then

Ly = sin (u -f-Aw) sin w

= sin w cosAM + cos u sinA% sin u

= cos u sinAM (1 cosAw) sin u;

. AV sinAM 1 cosAM .

whence - = cos M sm M.Au AM AM

Now let Ar and therefore AM approach zero. By (3), 42,

T sinAit ^ , , , A ^ , T . 1 cosAw A m. .

Lim - = 1, and, by (4), 42, Lim A= 0. Therefore

AM Aw<fo/~ = cos u.du

But by (8), 36, ^=Kdx dudx

and therefore -^ = cosw-^aa: dx

3

2. To find cos w, wo write

= sin u

rrn d d /Then -7- cos u = -~ sm

?23 tfo

/7T \?/

)

\2 /

'7T= coal

3. To find --- tan 7/, wo writedjc

, sin ?/

tan?< =cos ?/,

,, . aThen - tan u = ~

dx cos u

d , dcos u - sin ? sin u -~ cos ?*

eos4tt

VVH .

cos2w ^ y

f?.

;i

4. To find ~^~ cinu. wo writej?

, COS 7^1

ctn u = -isin u

rn ,d . d cosw

I hen ctn ?/ = .- -:d.r, ax Bin u

, d d .

sin u - cos w cos u -=- fiin M

126 TRIGONOMETBIC FUNCTIONS

5. To find sec u, we writeax

1 , ^1secw =- = (COSM) \cos u

Then sec w = - (cos )~* cos u (by (6), 36)

sum <2

j6. To find esc w, we write

aa;

esc u =- = (sin. u)~\sin M

3 7

Then esc u = (sin w)~2

sin u (by (6), 36)

-

Ex. 1. y = tan 2 a; tan3 a; = tan 2 x (tan a:)8.

^ = sec22 x-j- (2 a)- 2 (tana) tana;

dx dx^ ' v J dx

= 2 seo22 a; 2 tan a; sec2 a:.

Ex. 2. y = (2 sec*ar + 3 secaa;)

sin x.

= sin a:

^8sec8

*-! (see a) + 6 secx ~- (seca)J+ (2sec*a: + 3 sec2

a:)-^(sina;)

= sin x (8 sec*a; tan x + Q sec2 a; tana:) + (2 sec*a; + 3 secs

a:) cos x=

(1 cossa:) (8 sec6 a; + 6 sec8

a;) + (2 secs a: + 3 sec a;)

= 8 sec6 a: 3 sec ar.

EXERCISES

nd in each of the following cases :

y = 2tan|-

6. y = & sin85a! - sin5 .

sw.*2x. 7. ^rs

sa 5aj. 8 . .

SIMPLE HARMONIC MOTION 127

9. 2/= cos8 -2cos. 11. y

2 ., o * x ,o sec a* + ton as

10. y - ctn- + 2 ctn -. 12. y =

13. y = sin (2 a? + 1) cos (2 a; 1).

14. y = tan83 a; 3 tan 3 x + 9 ie.

15. y = see 2 a: tan 2 x.

16. y = ^ (3 cos62 8 5 cosB 2os).

17. sin 2 a + tan 3 ?/= 0.

18. asy + ctn xij= 0.

45. Simple harmonic motion. Let a particle of mass m move

in a straight line so that its distance s measured from a fixed

point in the line is given at any time t by the equation

s s=a G sin Si, (1)

where c and b are constants. We have for the velocity v and

the acceleration av ^ cbcos ^ (2)

a sss c62sin Ztf. (3)

When 0, 8 a and the particle is at (Fig. 54), When

t = -j^r*s =s c and the particle is at At where OA** o,

2 o

When t is between and v is positive and a is negative,>a y

so that the particle is moving from to A with decreasing speed.

When t is between and T v is i A25 o

, , ,. ,-, , ]?i<*. 64negative and a is negative, so that

the particle moves toward with increasing speed. When71"

==--, the particle is at 0.

8As * varies from ? to -^?? the particle moves with decreasing

b 26

speed from to J?, where OJ? =* o.

Finally, as varies from |^ to -, the particle moves back2o o

from J5 to (9 with increasing speed.

128 TRIGONOMETRIC FUNCTIONS

The motion is then repeated, and the particle oscillates between

B and A, the time required for a complete oscillation being, as

27Twe have seen, =- The motion of the particle is called simple

harmonic motion. The quantity c is called the amplitude, and tho2 7T

interval -= after which the motion repeats itself, is called

the period.

Since force is proportional to the mass times the acceleration,

the force F acting on the particle is given by the formula

F kma = kmcbssm bt= Jcmb*s.

This shows that the force is proportional to the distance s

from the point 0. The negative sign shows that the force pro-duces acceleration with a sign opposite to that of s, and there-

fore slows up the particle when it is moving away from arid

increases its speed when it moves toward 0. The force is there-

fore always directed toward and is an attracting force.

If, instead of equation (1), we write the equation

s = c8mb(t-tQ),-

(4)

the change amounts simply to altering the instant from which thetime is measured. For the value of s which corresponds to t t

in (1) corresponds to t = tl+t in (4). Hence (4) represents

simple harmonic motion of amplitude c and periodBut (4) may be written

*

s = G cos fa sin bt o sin bt cos bt,

which is the same as

s =A sin bt+B cos bt,

where A= c cos bt, B = c sin bt .

A and B may have any values in (5), for if A and B are given,we have, from the last two equations,

c=V^a+52, tanfa = --,

which determines G and t in (4).

SIMPLE HARMONIC MOTION 129

Therefore equation (5) also represents simple harmonic motion

2 7Twith amplitude VAz

+jBa and period -7

In particular, if in (5) A = and 33= c, we have

8 = G COS It. (6)

If in (4) we place t = - '

,it becomes

*j &

* =<! cos fi (-*), (7)

which differs from (6) only in the instant from which the time

is measured.

EXERCISES

1. A particle moves with constant speed v around a circle.

Prove that its projection on any diameter of the circle describes

simple harmonic motion

2. A point moves with simple harmonic motion of period 4 sec.

and amplitude 3 ft Find the equation of its motion.

3. Given the equation s = 5 sin 2 1 Find the tune of a completeoscillation and the amplitude of the swing.

4. Find at what time and place the speed is the greatest for the

motion defined by the equation s = G sin It Do the same for the

acceleration.

5. At what point in a simple harmonic motion is the velocity zero,

and at what point is the acceleration zero ?

6. The motion of a particle in a straight line is expressed by the

equation s = 5 2 cos3*. Express the velocity and the acceleration

in terms of s and show that the motion is simple harmonic.

7. A particle moving with a simple harmonic motion of amplitude5 ft has a velocity of 8 ft. per second when at a distance oj! 3 ft.

from its mean position. Find its period.

8. A particle moving with simple harmonic motion has a velocityof 6 ft. per second when at a distance of 8 ft. from its mean position,

and a velocity of 8 ft. per second when at a distance oC 6 ft. from its

mean position. Find its amplitude and its period.

9. A point moves with simple harmonic motion given by the

equation *= # sin et. Describe its motion,

130 TRIGONOMETRIC FUNCTIONS

46. Graphs of inverse trigonometric functions. The equation

x = sin y (1)

defines a relation between the quantities x and y which may be

stated by saying either that x 'is the sine of the angle y or that

the angle y has the sine x. When we wish to use the latter form

of expressing the relation, we write in place of equation (1)

the equation y = sm-^, (2)

where 1 is not to be understood as a negative exponent but as

part of a new symbol sin"1. To avoid the possible ambiguity

formula (2) is sometimes written

y = arc sin x.

Equations (1) and (2) have exactly the same meaning, and

the student should accustom himself to pass from one to the

other without difficulty. In equation (1) y is considered the

independent variable, while in (2) x is the independent variable.

Equation (2) then defines a function of x which is called the

anti-sine of x or the inverse sine of x. It will add to the clearness

of the student's thinking, however, if he will read equation (2)as

"y is the angle whose sine is x."

Similarly, if a = cos#, then y = GOS~'L

x; if x ioxiy, then

y tan"1a; ; and so on for the other trigonometric functions. We

get in this way the whole class of inverse trigonometric functions.

It is to be noticed that, from equation (2), y is not completelydetermined when x is given, since there is an infinite number

of angles with the same sine. For example, if # = - #=> >

STT 18 IT , _.. . t2

.

&

-~>ft

> etc. 1ms causes a certain amount of ambiguity in

using inverse trigonometric functions, but the ambiguity is re-

moved if the quadrant is known in which the angle y lies. Wehave the same sort of ambiguity when we pass from the equa-tion x = y* to the equation y = Va;, for if x is given, thereare two values of y.

To obtain the graph of the function expressed in (2) womay change (2) into the equivalent form (1) and proceed as

GRAPHS 131

in 43. In this way it is evident that the graphs of the inverse

trigonometric functions are the same as those of the direct func-

tions but differently placed with reference to the coordinate

axes. It is to be noticed particularly

that to any value of x corresponds an

infinite number of values of y.

X

FIG. 55 FIG 56

Ex. 1. y = sin- 1*

From this, x = sin y, and we may plot the graph by assuming values of

y and computing those of a (Fig. 55).

Ex. 2. y = tan- 1 *.

Then x == tan y, and the graph is as in Fig. 56.

EXERCISES

Plot the graphs of the following equations :

1. y tan- 1 2 x. 3. y = sin" 1(a: 1). 5. y =1+ cos" 1

:*;.

2. y =3 ctn-^cc. 4. y = tan- 1

(* + 1). 6. y = ^tan"1^

7. ?/= cos-^jc 2).

8. y sin-^S x 4. 1)- -

47. Differentiation of inverse trigonometric functions. Theformulas for the differentiation of the inverse trigonometricfunctions are as follows:

1. -7- sin""1^ = . -r- when sin""

1^ is in the first or theax vl wa <*x ,. , , , ,x u lourth quadrant ;

i diLi when sin-^f is in the second or

A/ "I M_ QI^ QiX ,t i t ix w the third quadrant.

132 TKIGONOMETKIC FUNCTIONS

2. - cos"1^ = , -7- when cos"

1** is m the first or the

^x vlw second quadrant;i (Jtjf= when cos" 1^ is m the third or the

VI - u2 dxfourth quadrant

d L. 1 duO ~z

dx 1+ wz ax

A d . _j 1 du4. ctn u=

3dx 1+

5. sec"1?* = when sec" 1^ is in the first or the

dx u^/u2-l dx

= when sec"1u is in the second or

uvu I ^ie fourtli quadrant.fj J. el/Mi

6 CSC"IM= . -7- when csc" 1^ is in the first or

x uvu 1|.jie third quadrant;

=, -r- when csc""

1^ is in the second or

The proofs of these formulas are as follows:

1. If y = sin~X

then sin y = u.

TT i P A A dy duHence, by 44, coS2,^

= -;

cfc 1 du,

whence -- --dx cos y dx

But cos y = Vl u2 when # is m the first or the fourth quad-rant, and cosy = Vl u2 when y is m the second or the third

quadrant.

2. If y = cos" 1

^,

then cos y u-

Hence _ sin^ =^;9 dx dx

whence *eta sin #

DIFFERENTIATION 183

But sin y Vl Ma when y is in the first or the second quad-rant, and am y = Vl u2 when y is in the third or the fourth

quadrant.

3. It' y tan' 1it,

then tan y = u.

it 2 d?t du1 1 ence sec u ~^~

;/yif>- /yVpdv*t Lt*v

, dy 1 duwhence -~ =

dx 1+ u dx

4. If yssctir1

!/,

then ctn ?/= w.

V T 41 (iV (/ ( V UV

, (??/ 1 duwhence ~ = = ; -7-

a 1+ w" are

5. If yssseo'X

then sec ?/= M.

1 1 ence sec y tan ?y-~= -r- ;ti*y* ft'V

. d'i/ 1 <??*

whence - =,

-=-*djc sec

;ytan y dx

But Rcoyss?/, and tan,y = V?^a --l when ?/ is in the first or

the third quadrant, and tan ?/:=-~-vV 1 when y is in the

second or the fourth quadrant.

M T$ /j/ rtiart '""'J'^f

then esc y = ?/,.

_ r . dy duHence esc y otn y j~

=sy ;

rvM/ \4iJu

whence -f-=

, -r-?a; esc y ofo&y dx

134 TRIGONOMETRIC FUNCTIONS

But cscy = w, and ctn^=Vw2 1 when y is in the first or

the third quadrant, and ctn# = Vw21 when y is in the

second or the fourth quadrant.If the quadrant in which an angle lies is not material in a

problem, it will be assumed to be in the first quadrant. This

applies particularly to formal exercises m differentiation.

Ex. 1. y = sin" 1Vl xz,where y is an acute angle

dv 1. .fLn. a>\|

^

L-(l-a2)

dxdx Vl -

This result may also be obtained by placing sin- 1Vl xz = cos" 1a:.

Ex. 2. y = sec" 1 V4a:a + 4ar + 2.

dy dx

dx V4 x* + 4 x + 2 V(4 a,-2 + 4 x + 2)

- 1

~~2a;2 + 2a,--

EXERCISES

Find -7^ in each of the following cases :

2. y = sin" 1 -.

_ -ift 3. y _ sin

g

i3a;

4. y = cos -1

11. y = COS~

2

5. y = tan~ 1

6. y = tan" 1Vcc2 2 x.

7. y = ctn- 1

^-

8. y = sec" 1 5 a.

9. y = csc~12aj.

, . a; + 610, y-tan->

13. y = tan" 1 Va3 - 1 +

16. y =

2\a a;/

18. y ss Vl aj9+a5003~

1 Vl wa

ANGULAR VELOCITY 135

48. Angular velocity. If a line OP (Fig. 57) is revolving in

a plane about one of its ends 0, and in a time t the line

OP has moved from an initial position OM to the position OP,the angle MOP = 6 denotes the amount of rotation. The rate

of change of with respect to t is

called the angular velocity of OP. The

angular velocity is commonly denoted

by the Greek letter <; so we have

the formula JQ

In accordance with 42 the angle 8

is taken in radians ; so that if t is in

seconds, the angular velocity is in

radians per second. By dividing by 2-Tr, the angular velocity

may be reduced to revolutions per second, since one revolution

is equivalent to TT radians.

A point Q on the line OP at a distance r from describes

a circle of radius r which intersects OM at A. If s is the lengthof the arc of the circle A Q measured from A, then, by 42,

s = rO. (2)

dsNow -r- is called the linear velocity of the point $, since it

dt

measures the rate at which s is described ; and from (2) and (1),

ds d0

showing that the farther the point Q is from. the greater is

its linear velocity.

Similarly, the angular acceleration, which is denoted by oc, is

denned by the relation , ,2 ,,Jaft> a o

This is connected with the linear acceleration -=-5 by the

formula

136 TRIGONOMETRIC FUNCTIONS

Ex 1. If a wheel revolves so that the angular velocity is given by the

formula u> = 8 1, how many revolutions will it make in the time from t = 2

to t = 5 ?

We take a spoke of the wheel as the line OP Then we have

dQ = 8 tdt

Hence the angle through which the wheel revolves in the given time is

B = C6

B tdt = [4 fif =100 - 16 = 84.1/2

The result is in radians. It may be reduced to revolutions by dividing by2 ir. The answer is 13.4 revolutions

Ex. 2. A particle traverses a circle at a uniform rate of n revolutions

a second Determine the motion of the projection of the particle on a

diameter of the circle

Let P (Fig 58) be the particle,

OX the diameter of the circle, and

M the projection of P on OX Then

x = a cos 6,

where a is the radius of the circle.

By hypothesis the angular velocity

of OP is 2 nir radians per second

Therefore ,

o> = = 2 mr;

dt

whence FIG. 68

If we consider that when t = 0, the particle is on OX, then (7 = 0.

Therefore /,

x = a cos 8 = a cos 2 rart = a cos <at.

The point M therefore describes a simple harmonic motion In fact,

simple harmonic motion is often defined in this way

EXERCISES

1. A flywheel 4ft in diameter makes 3 revolutions a second.

Find the components of velocity in feet per second of a point on the

rim when it is 6 m above the level of the center of the wheel.

2. A point on the rim of a flywheel of radius 5 ft which is 3 ft.

above the level of the center of the wheel has a horizontal componentof velocity of 100 ft per second Find the number of revolutions

per second of the wheel.

CYCLOID 1

3. If the horizontal and vertical projections of a point descr

simple harmonic motions given by the equations

x = 5 cos 3t, y = 5 sin 3 1,

show that the point moves in a circle and find its angular velocity

49. The cycloid. If a wheel rolls upon a straight line, ea

point of the rim describes a curve called a cycloid.

Let a wheel of radius a roll upon the axis of #, and let

(Fig. 59) be its center at any time of its motion, JV its point

NFIG. 59

contact with 6L3T, and P the point which describes the cycloi

Take as the origin of coordinates, 0, the point found by rollu

the wheel to the left until P meets OX.

Then ON= arc PN.

Draw MP and CN, each perpendicular to OX, PR parallel

OX, and connect C and P. Let

angle NGP <j>.

Then x= OM= 0-ZV- JfJV

=a<j> a sin <.

a a cos $

Hence the parametric representation ( 40) of the cycloid

a&=a(<f> sin<),

# = (! -cos<).

138 TRIGONOMETRIC FUNCTIONS

If the wheel revolves with a constant angular velocity to =-^

we have, by 40,

i) = a, C\. cos <f>")-*- = Q>* (1 cos <p),

?; = a sm<j> -If-

= am sw.<f> ',

at

whence va= a2

co2

(2-2 cos <)= 4 a26)2 sin2

1

v = 2 aco sm ^>

as an expression for the velocity in its path of a point on the

run of the wheel.

EXERCISESA

1. Prove that the slope of the cycloid at any point is ctn^-

2. Show that the straight line drawn from any point on the rim

of a rolling wheel perpendicular to the cycloid which that point is

describing goes through the lowest point of the rolling wheel.

3. Show that any point on the run of the wheel has a horizontal

component of velocity which is proportional to the vertical height of

the point

4. Show that the highest point of the rolling wheel moves twice

as fast as either of the two points whose distance from the ground as

half the radius of the wheel

5. Show that the vertical component of velocity is a maximum

when the point which describes the cycloid is on the level of the

center of the rolling wheel.

6. Show that a point on the spoke of a rolling wheel at a distance

b from the center describes a curve given by the equations

x = a<f> b sin <, y = a b cos<j>,

and find the velocity of the point in its path. The curve is called a

trochoid.

7. Find the slope of the trochoid and find the point at which the

curve is steepest.

8. Show that when a point on a spoke of a wheel describes a

trochoid, the average of the velocities of the point when in its highest

and lowest positions is equal to the linear velocity of the wheel

CUJftVATUKE 139

50. Curvature. If a point describes a curve, the change of

direction of its motion may be measured by the change of the

angle < ( 15).

For example, in the curve of Fig. 60, if AJ%= s and ./J^ = As,

and if fa and fa are the values of<j&

for the points Pl and Pzrespectively, then

</>2 ^ is the total change of direction of the

curve between J? and Py If Y

fafa= A<, expressed in

circular measure, the ratio

is the average changeAs

of direction per linear unit

of the arc PJ\. Regardingas a function of s, and

taking the limit of asAs

As approaches zero as a

limit, we have -f-> which is called the curvature of the curve atas

the point P. Hence the curvature of a curve is the rate of changeof the direction of the curve with respect to the length of the arc.

If is constant, the curvature is constant or uniform ; other-ds

wise the curvature is variable. Applying this definition to the

circle of Fig. 61, of which the Ycenter is C and the radius is a,

FIG CO

we have A<

As = a Arf>.

di--

P^CP^, and hence

Therefore - = -

-jAs a

Hence - and the circle is

as a

a curve of constant curvature equal

to the reciprocal of its radius.

The reciprocal of the curva-

ture is called the radius of cur-

vature and will be denoted by p. Through every point of a curve

we may pass a circle with its radius equal to p, which shall have

the same tangent as the curve at the point and shall lie on the

FIG. 61

140 TRIGONOMETRIC FUNCTIONS

same side of the tangent. Since the curvature of a circle is

uniform and equal to the reciprocal of its radius, the curvatures

of the curve and of the circle are the same, and the circle shows

the curvature of the curve in a manner similar to that in which

the tangent shows the direction of the curve. The circle is

called the circle of curvature.

From the definition of curvature it follows that

_ ds

If the equation of the curve is in rectangular coordinates,

ds

by (9), 36, p =~dx

To transform this expression further, we note that

,2 , 2 . ,2

whence, dividing by dx* and taking the square root, we have

Since<j>= tan- 1

, (by 15)

dx3

<dx ^ . fdy\

\dx,

Substituting, we have p = 2

x

d ydx*

In the above expression "for p there is an apparent ambiguity of

sign, on account of the radical sign. If only the numerical value

of p is required, a negative sign may be disregarded.

CUBVATUEE 141

3,2 y2Ex. 1. Find the radius of curvature of the ellipse + *j

= 1.

dy &aa:Here -f

= - -=-da; aay

Therefore P

Ex. 2. Find the radius of curvature of the cycloid ( 49).

We have -^ = a(l cos<) = 2asin22,a<p <s

d?/ . n A i^ = a sin

/>=s 2 a sm J cos ^

d<p it &

Therefore, by (9), 36,

f^ = ctn^.dx 2

and

EXERCISES

1. Find the radius of ourvature of the curve y* -fa a*.2222. Find the radius of curvature of the curve x* + y* =

3. Find the radius of curvature of the curve y taa-^a; 1)

at the point for which x 2

/ 7T\3

4. Show that the circle ( x -r- ) + y2= 1 is tangent to the curve

y = sin a; at the point for which x = -jr, and has the same radius of

curvature at that point.

B. Find the radius of curvature of the curve x = cos t, y = cos 2 1,

at the point for which t = 0,

6. Find the radius of curvature of the curve x = a cos < -{-

a< sin A, y =s a sin<]!> a<j! cos <.

7. Prove that the radius of curvature of the curve as = o.cos8<,

lias its greatest value when<j>= TT

142 TRIGONOMETRIC FUNCTIONS

51. Polar coordinates. So far we have determined the posi-

tion of a point in the plane by two distances, x and y. We may,however, use a distance and a direction, as follows :

Let (Fig. 62), called the origin, or pole, be a fixed point, and

let OM, called the initial line, be a fixed line. Take P any pointin the plane, and draw OP. Denote OP by r, and the angleMOPby 6 Then r and 6 are called the polar coor-

dinates of the point P(r, 0), and when givenwill completely determine P.

For example, the point (2, 15) is plotted

by laying off the angle MOP =15 and meas-

uring OP= 2.

OP, or r, is called the radius vector, and6 the veotorial angle, of P. These quantities may be either

positive or negative. A negative value of 6 is laid off in the

direction of the motion of the hands of a clock, a positive anglem the opposite direction. After the angle 6 has been constructed,

positive values of r are measured from along the terminal

line of 0, and negative values of r from along the backwardextension of the terminal line. It follows that the same point

may have more than one pair of coor-

dinates. Thus (2, 195), (2, -165),(-2, 15), and (- 2,

- 345) refer to

the same point. In practice it is usu-

ally convenient to restrict to positivevalues.

Plotting in polar coordinates is facili-

tated by using paper ruled as m Figs. 64and 65. The angle 6 is determined from

"

ElG 63the numbers at the ends of the straight

hnes, and the value of r is counted off on the concentric circles,

either toward or away from the number which indicates 6,

according as r is positive or negative.The relation between (r, 0) and (x, y) is found as follows :

Let the pole and the initial line OM of a system of polarcoordinates be at the same time the origin and the axis of # of a

system of rectangular coordinates. Let P (Fig, 63) be any point

POLAR COORDINATES 143

of the plane, (a;, #) its rectangular coordinates, and (r, 0) its

polar coordinates. Then, by the definition of the trigonometric

functions,XJ

Xcos 6 = -

sm = 2-.

Whence follows, on the one hand,

x = r cos

y = r sin i

and, on the other hand,

sin0 = cos =

(1)

(2)

By means of (1) a transformation can be made from rectangular

to polar coordinates, and by means of (2) from polai to rectangular

coordinates

When an equation is given in polar coordinates, the corre-

sponding curve may be plotted by giving to 6 convenient values,

computing the corre-

sponding values of r,

plotting the resulting

points, and drawing a

curve through them.

Ex. 1. r a cos $

a is a constant which

may be given any con-

venient value We maythen find from a table of

natural cosines the value

of r which correspondsto any value of 9. Byplotting the points cor-

J *

xesponding to values of 6 FIG 04

from to 90, we obtain

the arc ABCO (Fig. 64). Values of 6 from 90 to 180 give the arc ODEA.

Values of 6 from 180 to 270 give again the arc ABCO, and those fiom 270

to 860 give again the arc ODEA. Values of 6 greater than 360 can clearly

give no points not already found. The curve is a circle.

awMS

144 TRIGONOMETRIC FUNCTIONS

Ex. 2. r= a sin 35.

As 6 increases from to 30, r increases from to a;as 9 increases

from 30 to 60, r decreases from a to,the point (r, 0) traces out the loop

040 (Fig. 65), which is evidently symmetrical with respect to the radius

OA. As 6 increases from

60 to 90, r is negative

and decreases from to

a;as increases from

90 to 120, r increases from isoj

a to ;the point (r, 0)

traces out the loop OBOAs & increases from 120

to 180, the point (r, 6}

traces out the loop OCO.

Larger values of & give

points already found, since

sin 3 (180 + 5)= - sm 3 6.

The three loops are congiu-

ent, because sm,3 (60 + 0)=sin30 This curveis called

a rose of three leaves.

r = i a V2 cos 2 0.

Ex. 3. r2 =

Solving for r, we have

Hence, corresponding to any values of 9 which make cos 2 9 positive, there

will be two values of r numerically equal and opposite in sign, and two

corresponding points of the curve symmetrically situated with respectto the pole If values are assigned to 9 which make cos 2 9 negative, the

corresponding values of r will be

imaginary and there will be no

points on the curve.

Accordingly, as 9 increases| ^ | $f

from to 45, r decreases numer-

ically from aV2 to 0, and the

portions of the curve in the first

and the third quadrant are con-

structed (Fig 66) ;as 9 increases from 45 to 135, cos 2 9 is negative, and

there is no portion of the curve between the lines 9 = 45 and 9 = 135,

finally, as 9 increases from 135 to 180, r increases numerically from to

aV2, and the portions of the curve in the second and the fourth quadiantare constructed The curve is now complete, as we should only repeat thecurve already found if we assigned further values to 0, it is called thelemniscate.

FIG 06

GRAPHS 145

Ex. 4. The spiral of Archimedes,

r = a8.

In plotting, 6 is usually considered

in cncular measuie When &= 0, r =,

and as 6 mci eases, ? inci eases, so that

the cuive winds an infinite uiuubei of

times around the ongin while leced-

ing from it (Fig 67) In the flgme the

heavy line represents the poitiou of

the spiral coriespondmg to positive values of 0, and the dotted line the

portion coriespondmg to negative values of 6

EXERCISES

Plot the graphs of the following curvesa

1. r = a sm 6 9. r = a sin8 -

2. r = asm 2010. ?'

2 = a2 sin3. r = a cos 30

11. ?* =4. r = a sin-- 12. ,, = a (l

_ cos 20).

13. r = a(l+2cos20)5 - r=acos

2'

14. r =atan06. 7- = 3 cos + 5 15. r = a tan 2

7. ?' = 3 cos + 3.* 116 r = t

8. r = 3 cos + 2'

1 + cos

Find the points of intersection of the following pairs of curves :

17. r = 2 sm 0, ? = 2-x/S cos 0.

18. r2= a2 cos 0, ;-2 = a2

sin 2

19. r = 1 + sin 0, r = 2 sin

20. r2=a2sin0,

'2 =a2 sm30

Transform the following equations to polar coordinates .

21 . .r?/= 4. 23. xz + y*-2ay = 0.

22 . a;2 + f - 4 </a? - 4 ay = 24. (a* + 2/

2

)

2= aa(

a-Z/

2

)

Transform the following equations to rectangular coordinates :

25. r= ft sec 6 27. r = atan0.

26. r = 2 n cos 28. r = a cos 2

* The curve is called a cardioid

t The curve is a parabola with the ongui at the focus.

146 TRIGONOMETRIC FUNCTIONS

52. The differentials <?r, d9, ds, in polar coordinates. We have

seen, in 39, that the differential of arc in rectangular coordinates

is given by the equation

(1)

If we wish to change this to polar coordinates, we have to

placex = r cos 0, y r sin 6 ;

whence dx = cos 6dr r sin 6 d6,

dy = sin Qdr + r cos 6d6.

Substituting in (1), we have

ds*= drz+ r*d6z. (2)

This formula may be remembered by means of an"elemen-

tary triangle"

(Fig. 68), constructed as follows :

Let P be a point on a curve r =/(0), the coordinates of Pbeing (r, 0), where OP = r and MOP = 6. Let Q be increased

by an amount d0t thus determining another

point Q on the curve. From as a center

and with a radius equal to r, describe an

arc of a circle intersecting OQ in R so that

O.E = OP = r. Then, by 42, PR, = rdd. NowEQ is equal to Ar, and PQ is equal to As. -^ 6gWe shall mark them, however, as dr and da

respectively, and the formula (2) is then correctly obtained bytreating the triangle PQJR as a right triangle with straight-linesides. The fact is that the smaller the triangle becomes as Qapproaches P, the more nearly does it behave as a straight-line

triangle ; and in the limit, formula (2) is exactly true.

Other formulas may be read out of the triangle PQR. Let usdenote by i/r

the angle PQJR, which is the angle made by thecurve with any radius vector. Then, if we treat the triangle PQRas a straight-line right-angle triangle, we have the formulas :

.

(3)dr k J

DIFFERENTIALS 147

The above is not a proof of the formulas. To supply the

proof we need to go through a limit process, as follows:

We connect the points P and Q by a straight line (Fig. 69)and draw a straight line from P per-

pendicular to OQ meeting OQ at S.

Then the triangle PQS is a straight-

line right-angle triangle, and therefore

,

chord PQSP arcPg

arc PQ'

chord PQ'FIG. 69

Now angle POQ = A0, arc PQ = As,

and, from the right triangle OSP, SP = OP sin POQ = r sm A0.

Therefore

rsuiA* arcPr

Ag chord P<2 A0 As chord PQ

Now let A0 approach zero as a limit, so that Q approaches Palong the curve. The angle SQP approaches the angle OPT,

where PT is the tangent at P. At the same time ap-A/9 rlfi

proaches 1, by 42;

-approaches by definition ;

andT>/~)

^aic ''

approaches 1, by 39. In this figure we denote theCllOlXl JL ty

angle OPT by >/rand have, from (4),

. . dB ,CNsmf = r , (5)

which is the first of formulas (3). It is true that in Fig. 69 we

have denoted OPT by -^ and that in Fig. 68 ^ denotes OQP.But if we remember that the angle OQP approaches OPT as a

limit when Q approaches P, and that in using Fig. 68 to read off

the formulas (3) we are really anticipating this limit process, the

difference appears unessential.

The other formulas (3) may be obtained by a limit process

similar to the one just used, or they may be obtained more

148 TEIGONOMETEIC FUNCTIONS

quickly by combining (5) and (2). For, from (2) and (5),

we have

whence cos -\lr = . (6)ds

By dividing (5) by (6) we have

rdd /7NCOdr

In using (7) it may be convenient to write it in the form

tan i/r= , (8)

dr

d0

since the equation of tho curve is usually given in the formt?rt

r =/(#), and - is found by direct differentiation.cLu

Ex. Find the angle which the cmve 1 = a sin 4 makes with the radius

vector 6 = 80

Here ^ = 4 a cos 4 0. Therefore, fiom (8), tan ^ =" sin 4

^ = i tan 'I 6do ^ T 4 cos 4 6 4

Substituting 6 = 30, we have tan ^ = * tan 120 = - | V3 = - 4MOTherefore

j/r= 156 35'.

EXERCISES

1. Find the angle -which the curve r = a cos 3 makes with the

radius vector 6 = 45

2. Find the angle which the curve r = 2 + 3 cos makes with the

radius vector 6 = 90A

3. Find the angle which the curve r = a2 sina makes with the

initial line.

9 64. Show that for the curve r = a, sin8

5 ji/r= -

o 3

5. Show that the angle between the cardioid r = a(l cos 0) and

any radius vector is always half the angle between the radius vector

and the initial line.

GENEHAL EXERCISES 149

6. Show that the angle between the lemniscate r2 = 2 .

2 cos 2 9

TT

and any ladms vector is always plus twice the angle between the

radius vector and the initial line

7. Show that the curves ra= aa sin 2 6 and ?-2 = a2 cos 2 6 inter-

sect at right angles

GENERAL EXERCISES

Find the graphs of the following equations :

a 1

4- 1 * o / L^1. ?/

= 4sm - 5. y = Ssmlas + TT )

2. y = cos (2 x 3) 6. vya = tan x.

3 . y = tan- 7- y = 2cos2(o!-

2)

4. ys= Jsin2aj + ^8in8a5. 8.?/= 3cos3h +

^

Find -f^ in the following cases :

CtttXj

9. y = 2x

10. y = ,Jtan (3 K + 2) + J tan

8

(3 x + 2).

12. tan (x + ?/) + tan (x y)= 0.

13. y=3ctnfi

|+ 5ctn8

|.21. y =

14. ?/= csca4 x + 2 ctn 4 x 22. ?/

=

16. ?/= sma4 x cos*2 a;

23> V cos~

a.a_|_ 4>

- cos8 - 2 cos ~ 24. v = ctn" 1 Vsc2 2x.17. ?/=

218. ?y=atan

82a! ^tan2C+03. 25. 2/

= csc" 1

1t SC "y" X

19. y = sin

20. ?/ = cos" 1

3

28. A particle moves in a straight line so that s = 6

Show that the motion is simple harmonits and find the center about

which, the particle oscillates and the- amplitude of the motion.

150 TRIGONOMETRIC FUNCTIONS20

?/

29. A particle moves on the ellipse i + ^ = 1 so that its projeq-

tion upon OX describes simple harmonic motion given by x = a cos kt.

Show that its projection upon OF also describes simple harmonic

motion and find the velocity of the particle in its path.7T

30. A particle moving with simple harmonic motion of period -r

has a velocity of 9 ft per second when at a distance of 2 ft. from its

mean position. Find the amplitude of the motion.

31. A particle moves according to the equation s = 4 sin \t +5 cos t. Show that the motion is simple harmonic and find the

amplitude of the swing and the time at which the particle passes

through its mean position.

32. Find the radius of curvature of the curve y = x sin- at the

2 x

point for which x =7T

33. Find the radius of curvature of the curve ?/ = at the7!!

point for which x = TT

34. Find the radius of curvature of the curve y a sin'1' V,2

.r*

CL

at the point for which x = r

35. Findthe radius of curvature of the curve x => a cos <, y = I sin<{>,

ITat the point for which

<j>= -r-

Plot the graphs of the following curves .

a

36. r!=a2

sm|-41. r =1-20.42. ra =

37. ra = a2 sin40 .. ,43. r*=

38. r =a(l-sin0). .v '44. r =l + sm s .

39. r =a(l+cos20). ^

3040. r = a(l + 2 sin 0). 45. r =* 1 + sin

-7p

Find the points of intersection of the following pairs of curves ;

46. r*=3cos20, ?-s = 2cosa

0.

47. r BS a cos 0, r2 BB a2 sin 2 0.

48. r 2 sin 0, 7-' B* 4 sin 2 0.

49. r ~a(l + sin 20), r2=? 4 a'siii 2 0.

GENERAL EXERCISES 151

Transform the following curves to polar coordinates :

50. = "

Transform the following curves to ^-coordinates

52 ?'2 =2aa sm20. 53. r = a(l cos0)

54. Find the angle at which the curve r = 3 + sin 2 6 meets the

circle r 3

55. Find the angle of intersection of the two curves r = 2 sin 6

and r2 = 4 sin 2

56. Find the angle of intersection of the curves r = a cos and

r = a sin 2 B.

57. If a ball is fired from a gun with the initial velocity v,

it

as?describes a path the equation of which is i/

= x tan a'

r i J

where a is the angle of elevation of the gun and OX is horizontal

What is the value of a when the horizontal range is greatest ?

58. In measuring an electric current by means of a tangent galva-

nometer, the percentage of error due to a small error in reading is

proportional to tan x + ctn x. For what value of x will this percent-

age of error be least ?

59. A tablet 8 ft high is placed on a wall so that the bottom of

the tablet is 29 ft. from the ground. How far from the wall should

a person stand in order that he may see the tablet to best advantage

(that is, that the angle between the lines from his eye to the top and

to the bottom of the tablet should be the greatest), assuming that

his eye is 5 ft. from the ground ?

60. One side of a triangle is 12 ft. and the opposite angle is 36

Find the other angles of the triangle when its area is a maximum

61. Above the center of a round table of radius 2 ft is a hanging

lamp. Plow far should the lamp be above the table in order that

the edge of the table may be most brilliantly lighted, given that

the illumination varies inversely as the square of the distance and

directly as the cosine of the angle of incidence ?

62. A weight P is dragged along the ground by a force F. If

the coefficient of friction is k, in what direction should the force be

applied to produce the best result ?

L52 TRIGONOMETRIC FUNCTIONS

63. An open gutter is to be constructed of boards in such a way,hat the bottom and sides, measuied on the inside, are to be each

3 in wide and both sides are to have the same slope How wide

should the gutter be across the top in oider that its capacity mayje as great as possible

?

64. A steel girdei 27 ft long is to be moved on rollers along a

Dassageway and into a corridor 8 ft. in width at right angles to the

Dassageway. If the horizontal width of the girder is neglected, how

vide must the passageway be in order that the girder may go around

-he corner 9

65. Two particles are moving in the same straight line so that

iheir distances from a fixed point are respectively x = a cos kt and

e' = acosud -f qOj& and a being constants Find the greatest

hstance between them

66. Show that for any curve in polar coordinates the maximumLiid the minimum values of r occui in general when the radius vector

s perpendicular to the curve.

67. Two men aie at one end of the diameter of a circle of 40 ydadius. One goes directly toward the center of the circle at the

miform rate of 6 ft. pei second, and the other goes around the

.ircumference at the rate of 2 TT ft per second How fast are they

eparatmg at the end of 10 sec. 'f

68. Given that two sides and the included angle of a triangle are

> ft,10 ft

,and 30 respectively, and are changing at the rates of

- ft,

3 ft,and 12 per second respectively, what is the area of the

riangle and how fast is it changing ?

69. A revolving light in a lighthouse mi offshore makes one

evolution a minute If the line of the shore is a straight line, howast is the ray of light moving along the shore when it passes a

ioint one mile from the point nearest to- the lighthouse?

70. BC is a rod a feet long, connected with a piston rod at C, andt B with a crank AB, b feet long, revolving about A. Find C's

elocity in terms of All's angular velocity.

71. At any time t the coordinates of a point moving in the ajy-plane

re x = 2 3 cos t, y = 3 + 2 sin t Find its path and its velocity in

,s path. At what points will it have a maximum speed?

72 . At any time t the coordinates of a moving point are x= 2 sec 3 1,

= 4 tan 3 1. Find the equation of its path and its velocity in its path.

GENERAL EXERCISES 153

73. The parametric equations of the. path of a moving particle are

O3 = 2cos8

<, 2/=2sin8

< If the angle < increases at the rate of

2 radians per second, find the velocity of the particle in its path

74. A particle moves along the curve y = smo3 so that the

a-component of its velocity has always the constant value a Find

the velocity of the particle along the curve and determine the points

of the curve at which the particle is moving fastest arid those at

which it is moving most slowly

75. Find the angle of intersection of the curves y = smx and

y = cos x

76. Find the angle of intersection of the curves y = sma; and

/ , ^2/= sm(a: + ^

)

77. Find the angle of intersection of the curves y since and

y = cos 2 x between the lines x = and ce = 2 TT.

78. Find the points of intersection of the curves T/= since and

y = sin 3 x between the lines x = and x = TT Determine the angles

at the points of intersection.

79. Find all the points of intersection of the curves y = cos x and

y = sin 2 x which, lie between the lines x = and x = 2 TT, and

determine the angles of intersection at each of the points found.

CHAPTER VI

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

53. The exponential function. The equation

/=

*,

where a is any constant, defines y as a function of x called the

exponential function.

If x = n, an integer, y is determined by raising a, to the nth

power by multiplication.

If #=*- a positive fraction, y is the jth root of the plli

power of a.

If x is a positive irrational number, the approximate value of

y may be obtained by expressing oc approximately as a fraction.

If x = 0, y = a=l. If x =- m, y = a~ m =The graph of the function is readily found.

Ex. Find the gi aph of y = (1 5y. By giving convenient values to x

we obtain the curve shown in Fig 70 To determine the shape of the

curve at the extreme left, we place a equal to a large negative number,

say x = - 100 Then y = (1.5)-=^ J

,

which is very small It is obvious that the

larger numencally the negative value of x

becomes, the smaller y becomes, so that the

curve appi caches asymptotically the negative _.

portion of the a:-axis. OOn the other hand, if # is a large positive

number, y is also large. FIG. 70

54. The logarithm. If a numberN may be obtained by placingan exponent L on another number a and computing the result,

then L is said to be the logarithm of .2V to the base a. That is, if

N= d>, (1)then L = loga JV. (2)

154

LOGARITHMS 155

Formulas (1) and (2) are simply two different ways of ex-

pressing the same fact as to the relation of ,JV" and L, and the

student should accustom himself to pass from one to the other

as convenience may demand.

From these formulas follow easily the fundamental propertiesof logarithms; namely,

M= log

=logaN, (3)

logal=0,

loga- = -log JV.

Theoretically any number, except or 1, may be used as

the base of a system of logarithms. Practically there are only

two numbers so used. The first is the number 10, the use of

which as a base gives the common system of logarithms, which

are the most convenient for calculations and are used almost

exclusively in trigonometry.Another number, however, is more convenient in theoretical

discussions, since it gives simpler formulas. This number is

denoted by the letter e and is expressed by the infinite series

where 21=1x2, 31=1x2x8, 41=1x2x3x4, etc.

Computing the above series to seven decimal places, we have

e = 2,7182818....

An important property of this number, which is necessary in

finding the derivative of a logarithm, is that

156 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

To check this arithmetically we may take successive small

values of h and make the following computation :

When &=.!, (1+ A)*= (1.1)10= 2.59374.

When h =.01, (1+ A)*=(1.01)100= 2.70481.

When h = .001, (1 + hy= (1.001)1000= 2.71692.

When h =.0001, (1+ hy= (1.0001)10000= 2.71815.

i

Working algebraically, we expand (1+A)Aby the binomial

theorem, obtaining ... . _._. ...- 11 if i~1 V - 2

A*

2! 8!

where JS represents the sum of all terms involving A, 7ia,

7i8, etc.

Now it may be shown by advanced methods that as h approaches

zero, B, also approaches zero ; so that

When the number e is used as the base of a system of loga-

rithms, the logarithms are called natural logarithms, or Napierian

logarithms. We shall denote a natural logarithm by the symbol

In*; thus,

N~*>(4)

then L = InN.

Tables of natural logarithms exist, and should be used if

possible. In case such a table is not available, the student

* This notation is generally used by engineers. The student should fenow

that the abbreviation "log" is used by many authors to denote the natural

logarithm. In this book "log

"is used for the logarithm to the base 10.

LOGAEITHMS 157

may find the natural logarithm by use of a table of commonlogarithms, as follows-

Let it be required to find In 213.

If x = In 213,

then, by (4), 213 = ev

;

whence, by (3), log 213 = a? log e,

or log 213 2.3284

Iog2.7183 0.4343

Certain graphs involving the number ye are important and are shown in the

examples.o

Ex. 1. y = In or.

Giving x positive values and finding y, weobtain Fig 71.

Ex. 2. y = e-*? I FIG. 71

The curve (Fig 72) is symmetrical with respect to OY and is alwaysabove OX. When a; = 0, y = 1 As a: increases numerically, y decreases,

approaching zero. Hence OX is an asymptote ^

FIG. 72 FIG. 73

Ex.3. y =

This is the curve (Fig. 73) made by a cord or a chain held at the ends

and allowed to hang freely. It is called the catenary.

Ex. 4. y e~ *" sin Ix.

The values of y may be computed by multiplying the ordmates of the

curve y = <s~ oa! by the values of sin bx for the corresponding abscissas. Since

the value of sinZw oscillates between 1 and -1, the values of e-

158 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

cannot exceed those of e-** Hence the graph lies in the portion of the

plane between the curves y = e~ axy

and y = e~ ax

. When a: is a mul-

tiple of -, y is zero. The graph

therefore crosses the axis of x an

infinite number of times Fig. 74

shows the graph when a= 1, b= 2 TT

Ex. 5. y ex

When x approaches zeio, being

positive, y increases without limit -1''

When x approaches zero, being neg-

ative, y approaches zeio,for example, when

x = Ttnjff' y = e1000' and when x ~ ~~ "W11 = e

-1000 = - The function is thereforeJglOOO

discontinuous for x

The line y - 1 is an asymptote (Fig. 75),

for as x increases without limit, being posi-

tive or negative, approaches 0, and y

appioaches 1.

Ex. 6. r = ea .

The use of r and 6 indicates that we are

using polar coordinates.

When 6 = 0, r 1. As increases, r in-

creases, and the curvewinds around the origin

at increasing distances from it (Fig 76).

When 6 is negative and increasing numer-

ically without limit, r approaches zero.

Hence the curve winds an infinite number

of times aiound the origin, continually ap-

proaching it The dotted line in the figure

corresponds to negative values of 6

The curve is called the logarithmic spiral

EXERCISES

Plot the graphs of the following equations

! y = ()* 5 - y = x&x-

?a*

74

X

75

M

3. y=B

4. y

7. y = log2o:.

1

'a:

8. =

FIG. 76

9. y = log sin x.

10. y = log tan ar.

11. y = e~ 2arsin4ai.

12. y = e~ x cos 3 a;.

13. r=e-* e.

EMPIKICAL EQUATIONS 159

55. Certain empirical equations. If x and y are two related

quantities which are connected by a given equation, we mayplot the corresponding curve on a system of ^-coordinates, and

every point of this curve determines corresponding values of

x and y.

Conversely, let x and y be two related quantities of whichsome corresponding pairs of values have been determined, andlet it be desired to find by means of these data an equation con-

necting x and y in general. On this basis alone the problemcannot be solved exactly. The best we can do is to assume that

the desired equation is of a certain form and then endeavor to

adjust the constants in the equation in such a way that it fits

the data as nearly as possible. We may proceed as follows :

Plot the points corresponding to the known values of x and y.

The simplest case is that in which the plotted points appear to

lie on a straight line or nearly so. In that case it is assumed

that the required relation may be put in the form

y = mx + *, (1)

where m and I are constants to be determined to fit the data.

The next step is to draw a straight line so that the plotted

points either lie on it or are close to it and about evenly dis-

tributed on both sides of it. The equation of this line may be

found by means of two points on it, which may be either two

points determined by the original data or any other two points

on the line.

The resulting equation is called an empirical equation and

expresses approximately the general relation between x and y.

In fact, more than one such equation may be derived from the

same data, and the choice of the best equation depends on the

judgment and experience of the worker.

Ex. 1. Corresponding values of two related quantities x and y are given

by the following table :

ai 1 2 4 6 10

y 1.8 2.2 2.9 3.9 6.1

Find the empiucal equation connecting them.

160 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

We plot the points (a;, y) and draw the straight line, as shown, in

Fig. 77. The straight line is seen to pass through the points (0, 1) and

(2, 2). Its equation is thereforeT7"

?/= 5 x + 1,

which is the required equation

111 many cases, however,the plotted points will not

appear to lie on or near a

straight line. We shall con-

sider here only two of these ~o i

cases, which are closely con- -

FIG ?7nected with the case just

considered. They are the cases in which it may be anticipated

from previous experience that the required relation is either

of the formy=db*, (2)

where a and b are constants, or of the form

y=atf, (3)

where a and n are constants.

Both of these cases may be brought directly under the first

case by taking the logarithm of the equation as written. Equa-tion (2) then becomes

log y = log a, + x log 6. (4)

As log a and log I are constants, if we denote log y by y',

(4) assumes the form (1) in x and y\ and we have only to

plot the points (#, y'~) on an o^'-system of axes and determine

a straight line by means of them. The transformation from. (4)back to (2) is easy, as shown in Ex. 2.

Taking the logarithm of (3), we have

log y = log a + n log x. (5)

If we denote logy by y' and logo; by x\ (5) assumes the form

(1) in. a/ and /, since log a and n are constants. Accordinglywe plot the points (V, y

r

~)on an a/?/-system of axes, determine

the corresponding straight line, and then transform back to (8),as shown in Ex. 3.

EMPIKICAL EQUATIONS 161

Ex. 2. Corresponding values of two related quantities x and y are given

by the following table .

a; 8 10 12 14 16 18 20

y 3.2 40 73 98 152 240 364

Find an empirical equation of the form y = dbf*

Taking the logarithm of the equation y = ob*, and denoting log y by y',

we have'

if= log a + x log i.

Determining the loganthm of each of the given values of y, we form a

table of corresponding values of x and y', as follows :

x 8 10 12 14 16 18 20

if = logy 5051 6028 8633 9912 1.1818 1.3909 15611

We choose a large-scale plotting-paper, assume on the y'-axis a scale

four times as laige as that on the

or-axis, plot the points (x, /), and

20

J,point. Its equation is 10 >

tour times as laige as tnat on tne *

or-axis, plot the points (x, /), and i

j

draw the straight line (Fig 78) 20 I

H-iivM-in-li +.Vo -fifaf. o.nrl tliA HTvKh i ethrough the first and the sixth 15

' = .08858 x - .20354.5 m

2 4 6 8 10 12 14 16 18 20 *

Therefore log a = 20354 = FlG ?g9 7965 - 10, whence a = .626

;and

log & = .08858, whence & = 1 22. Substituting these values in the assumed

equation, we have

as the required empirical equation The result may be tested by substitxit-

ing the given values of x in the equation. The computed values of y will

Le found to agree fairly well with the given, values.

Ex. 3. Corresponding values of pressure and volume taken from an

indicator card of an air-compressor are as follows:

p 18 21 205 33.5 44 62

v .035 .656 .476 .897 .321 .243

Find the relation between them in the tormpv"~

c,

162 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Writing the assumed relation, in. the form p = cv~ n and taking the

logarithms of both sides of the equation, we have

logp=n log v + log c,

or

where y = log jtjjx = log v, and 6 = log c.

The corresponding values of x and y are

- 4012 -.4935 - 6144

1 5250 1 6435 1 7994

x = logu 1972

y = logp 1 2553

- 2549 - .3233

1 3222 1 4232

For convenience we assume on the ar-axis a scale twice as large as that on

the y-axis, plot the points (x, ?y), and diaw the straight line as shown in

Fig 79 The construction should

be made on large-scale plotting-

paper. The line is seen to pass

through the points (- 05, 1 075)and ( .46, 1.6) Its equation is

therefore

y=- 128 x +101

Hence n 1 28, log c = 1 01,

c = 10 2, and the required rela-

tion between p and v is

pvi as = 10 2

) -.55-50-lt5-W-SS-SO~25-0-15 10- OS

FIG. 79

EXERCISES

1. Show that the following points lie approximately on a straight

line, and find its equation :

4

21 46

13

7

20

12

22

12 9

25

14,5

30

18.2

2. For a galvanometer the deflection D, measured in millimeters

on a proper scale, and the current /, measured in microamperes, are

determined in a series of readings as follows :

D 291 48.2 72.7 92.0 118.0 140.0 165.0 109.0

I 0.0493 0.0821 0123 0.154 0197 0234 0.274 0328

Find an empirical law connecting D and /.

EMPIRICAL EQUATIONS '

-, 163

3. Corresponding values of two related quantities x and y are

given in the following table

x 01 03 0.5 0.7 0.9 1.1 1 3 1.5

y 3316 4050 4046 0.6041 0.7379 9013 1 1008 1.3445

Find an empirical equation connecting x and y in the form y = ab*.

4. In a certain chemical reaction the concentration c of sodmmacetate produced at the end of the stated number of minutes t is

as follows :

* 1 2 8 4 5

c 00837 00700 00586 00492 00410

Find an empirical equation connecting c and t in the form c = atf

5. The deflection a of a loaded beam with a constant load is

found for various lengths I as follows

Z 1000 900 800 700 600

a 7.14 5.22 3.64 242 1.50

Find an empirical equation connecting a and I in the form a = nlf1

6. The relation between the pressure j and the volume v of a gasis found experimentally as follows :

p 20 23.5 31 42 59 78

v 0019 0.540 0.442 0358 0277 0219

Find an empirical equation connecting p and v in the formpvn = c.

56. Differentiation. The formulas for the differentiation of

the exponential and the logarithmic functions are as follows,

where, as usual, u represents any function which can be differen-

tiated with respect to #, In means the Napierian logarithm, and

a is any constant:d , loga e du ,1 .

-r-loga tt = fia--r (1)das u da

d , 1 du ^o^--- In u = -3-, (2)

aa? u ax

164 EXPONENTIAL AND LOaAEITHMIC FUNCTIONS

dx"~- ~dx (3)

d u B dudx dx' ^ -'

The proofs of these formulas are as follows:

.j

To find -logaM place y = log M.

Then, if u is given an increment AM, y receives an increment

Ay, where

Ay = loga(M+ AM) - log M

Aw, A,AM\A

)'

u /

the transformations being made by (3), 54.

ThenAM

Now, as AM approaches zero the fraction may be takenA of 54.

uas A of 54.

u

Hence Lim fl 4- Y"= e.

A-0\ M /

and

2. If y = hi u, the base a of the previous formula is e ; and

since log e=l, we have

dy _ 1 du

dx u dx

DIFFERENTIATION 165

3. If # = ",

we have In y = In au = u In a.

Hence, by formula (2),

1 dy , du-- = lna ;

y ax ax

. dy ul duwhence ~ = a In a -

ax ax

4. If y = ew the previous formula becomes

^ = e ^.dx dx

Ex. 1. y = In (a;2 -4 a + 5).

dy _ 2 x 4

da; a;2 4 a; + 5

Ex.2. y = '-

Ex. 3. y = e-

^ = cos bx (e~dx dx

EXERCISES

Find in each of the following cases :

dx-- I~sin2aj

2. y*=

3. JJ^-..

4. y = asln""X 12. y = e" a>e sin 3 cc.

5. /.

y^ln6. y = lnV2a;a +6a; + 9.

*

, a; - 3 14. y - 68a (9 a2 - 6ai + 2).

7 . w ==-J In-

5x + 3_ 16. y e*x(2 sin a: - cos

).

8. = ln(flj + Vaj8 + 4). a: 9

9. 3/=

17. y = sec x tan 03 + In (sec a? -f- tan *).

. VaTTl-118. v In

166 EXPONENTIAL AND 'LOGARITHMIC FUNCTIONS

57. The compound-interest law. An important use of the ex-

ponential function occurs in. the problem to determine a,function

whose rate of change is proportional to the value of the function.

If y is such a function of #, it must satisfy the equation

where ~k is a constant called the proportionality factor.

We may write equation (1) in the form

ldV-7f .r /fc ,

ydx

whence, by a very obvious reversal of formula (2), 56, we have

In y = kx 4- (7,

where C is the constant of integration (18).From this, by (1) and (2), 54,

Finally we place ec=A, where A may be any constant, since

C is any constant, and have as a final result

(2)

The constants A and Jc must be determined by other condi-

tions of a particular problem, as was done in 18.

The law of change here discussed is often called the compound-interest law, because of its occurrence in the following problem :

Ex. Let a sum of money P be put at interest at the rate of r% per annum.A

The interest gained in a time Ai is -Pr^ Ai, where A* is expressed in

years. But the interest is an increment of the principal P, so that we have

In ordinary compound interest the interest is computed for a certain

interval (usually one-half year), the principal lemaming constant duringthat interval The interest at the end of the half year is then added to the

principal to make a new principal on which interest is computed for th.9

COMPOUND-INTEREST LAW 167

next half year. The principal P therefore changes abruptly at the end of

each half year.

Let us now suppose that the principal changes continuously ; that is,

that any amount of interest theoretically eained, in no matter how small

a time, is immediately added to the puncipal. The average rate of changeof the principal in the peuod Ai is, fiorn 11,

* = lL mAi 100 ( }

To obtain the rate of change we must let A< approach zero in equation

(1), and havedp ?

,

From this, as in the text, we haver

P=Aelo. (2)

To make the problem conci ete, suppose the original principal were $100and the rate 4%, and we ask what would be the principal at the end of 14 yr.

We know that when t = 0, P = 100. Substituting these values in (2), wehave A = 100, so that (2) becomes

JL, 1P = 100e100 =100e20

Placing now t = 14, we have to compute P = 100 ei e The value of Pmay be taken from a table if the student has access to tables of powers of e

In case a table of common logarithms is alone available, P may be found

by first taking the logarithm of both sides of the last equation. Thus

logP = loglQO + it loge = 24053;

whence P = $254, approximately

EXERCISES

1. The rate, of change of y with respect to x is always equal to

\ y, and when x = 0, y = 5. Find the law connecting y and x,

2. The rate of change of y with lespect to x is always 0.01 times?/,

and when x = 10, y = 50. Find the law connecting y and x.

3. The rate of change of y with respect to x is proportional to y,

When x = 0, y = 7, and when x = 2, y = 14 Find the law connect-

ing y and x.

4. The sum of $100 is put at interest at the rate of 5% per annumunder the condition that the interest shall be compounded at each

instant of time. How much will it amount to in 40 yr.?

168 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

6. At a certain date the population of a town is 10,000. Forty

years later it is 25,000. If the population increases at a rate which

is always proportional to the population at the time, find a general

expression for the population at any time t.

6. In a chemical reaction the rate of change of concentration of

a substance is proportional to the concentration at any time. If the

concentration is y^ when t = 0, and is T T when t = 6, find the law

connecting the concentration and the time.

7. A rotating wheel is slowing down in such a manner that the

angular acceleration is proportional to the angular velocity. If the

angular velocity at the beginning of the slowing down is 100 revolu-

tions per second, and in 1 min. it is cut down to 50 revolutions per

second, how long will it take to reduce the velocity to 25 revolutions

per second ?

GENERAL EXERCISES

Plot the graphs of the following equations :

i i

1 - y = ($)"* 4 - y el~ x

- ? y e?.

2. y = e1-*. 5. y = (e*+ -*). 8. y = aser*.

3. y = e SGOSX. 6. y=~

9. y=*y?&-*.Je'+e"* J

10. For a copper-nickel thermocouple the relation between the

temperature t in degrees and the thermoelectric power p in micro-volts is given by the following table :

t 50 100 160 200

P 24 25 26 26.9 27.5

Find an empirical law connecting t and p.

11. The safe loads in thousands of pounds for beams of the samecross section but of various lengths in feet are found as follows :

Length 10 11 12 18 14 16

Load 123.6 121.5 111.8 107.2 1018 90.4

Find an empirical equation connecting the data.

GENERAL EXERCISES 169

12. In the following table s denotes the distance of a movingbody from a fixed point in its path at time t

t I 2 4678s 10 4 6400 0.1024 0410 0164

Find an empirical equation connecting s and t in the form s = ab*.

13. In the following table c denotes the chemical concentration of

a substance at the time t .

t 2 4 6 8 10

c 00060 00048 0.0033 0.0023 00016

Find an empirical equation connecting c and t in the form c = ah*.

14. The relation between the length I (in millimeters) and the

time t (in seconds) of a swinging pendulum is found as follows :

I 634 805 90.4 1013 107.3 1406

t 0.806 0892 0.960 1010 1038 1.198

Find an empirical equation connecting I and t in the form t Tdn.

15. For a dynamometer the relation "between the deflection 8,

2irwhen the unit 6 = -r-rri and the current I, measured in amperes, is

as follows :

6 40 86 120 160 201 240 280 320 362

I 0.147 0215 0.252 0293 0.329 0360 0.390 0.417 0442

Find an empirical equation connecting J and 6 in the form I k$"

16. In a chemical experiment the relation between the concen-

tration y of undissociated hydrochloric acid and the concentration x

of hydrogen ions is shown in the table

x 1.68 1.22 784 426 0.092 047 0096 0.0049

y 1 82 676 216 074 0085 0.00815 0.00036 00014

Find an empirical equation connecting the two quantities m the

form y = kxn.

170 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

17. Assuming Boyle's law, pv = c, determine o graphically from

the following pairs of observed values :

39.92 42.17 45.80 48 62 51 80 (J0.47 CMS 1)7

40.37 38.32 85.32 33.29 31.22 2080

Tind-J-

in each of the following cases :

CCSs

. Bx-218. =

19. y = In sin a-.

e _ e-

20. ?/= tan" 1

&

21. y = In (2as + V4 aja

l) + 2a osc"^ a;.

_a22. y = xz

ex

.

23. = l

24. y = I tan2oas + In cos ax.

25 . y = x tan- 1^ $ In (1 + a;2

).

26. A substance of amount x is being decomposed at a rate which

is proportional to a;. If x = 3 12 when t = 0, and cc == 1.30 whunt = 40 min., find the value of x when t = 1 hr.

27. A substance is being transformed into another at a rate whiith

is proportional to the amount of the substance still imtranafovnuid,

If the amount is 50 when t = 0, arid 15.6 when t = 4 hr., lind how

long it will be before y^ of the original substance will rumain.

28. According to Newton's law the rate at which tho temperatureof a body cools in air is proportional to the difference botwoou tho

temperature of the body and that of the air. If the tempwraturo of

the air is kept at 60, and the body cools from 130 to 120" in 300 sue.,

when will its temperature be 100 ?

29. Assuming that the rate of change of atmospheric prnssuro pat a distance h above the surface of the earth is proportional to tho

pressure, and that the pressure at sea level is 14.7 Ib. per square inch

and at a distance of 1COO ft. above soa level is 18.8 Ib. per squareinch, find the law connecting^ and h.

GENERAL EXERCISES 171

30. Prove that the curve y = e~Zx sm Sec is tangent to the curve

y = e- 2* at any point common to the two curves.

31. At any time t the coordinates of a point moving in a planeare x = a~'

2t cos 21, y o~'2t sin 2 1. Find the velocity of the point at

any time t Find the rate at which the distance of the point from

the origin is decreasing. Prove that the path of the point is a loga-

rithmic spiral.

32. Show that tho logarithmic spiral r = e"9 cuts all radius vectors

at a constant angle

33. Find the radius of curvature of the curve y = e~Za sin 2x at

7Tthe point for which x =

ft ( 5

34. Show that tho catenary y n^+e n and the parabola1

^

y = a -f--

a;3 have tho same slope and the same curvature at their

i Q>

common point.

35 . Find the radius of curvature of the curve x =s et sin t, y = e* cos t.

36. Show that the product of the radii of curvature of the curveX

y =s ae " at the two points for which x = a is a?(e + e" 1

)8.

37. Find the radius of curvature of the curve ?/= In a; and its

least value.

38. Find the radius of curvature of the curve y = er cosaj at the

7T

point for which x = -=

\n)

CHAPTER VII

SERIES

58. Power series. The expression

afl+ a

:x + ajK?+ a$+ ajf-\

----, (1)

where aQl a^ a

z , are constants, is called a power series in x.

The terms of the series may be unlimited in number, in which

case we have an infinite series, or the series may terminate after

a finite number of terms, in which case it reduces to a polynomial.

If the series (1) is an infinite series, it is said to converge

for a definite value of x when the sum of the first n terms

approaches a limit as n increases indefinitely.

Infinite series may arise through the use of elementary opera-

tions. Thus, if we divide 1 by 1 a; in the ordinary manner,

we obtain the quotient

and we may write

LX (2)^ '

Similarly, if we extract the square root of 1 + x by the rule

taught in elementary algebra, arranging the work as follows;

l+x

2 + -

172

POWER SERIES 173

the operation may be continued indefinitely. We may write

The results (2) and (3) are useful only for values of x for

which the series in each case converges. When that happensthe more terms we take of the series, the more nearly is their

sum equal to the function on the left of the equation, and in

that sense the function is equal to the series. For example, the

series (2) is a geometric progression which is known to con-

verge when x is a positive or negative number numerically less

than 1. If we place x = % in (2), we have

"-

which is true in the sense that the limit of the sum of the terms

on the right is f . If, however, we place x 3 in (2), we have

which is false. A reason for this difference may be seen by

considering the remainder in the division which produced (2)

but which is neglected in writing the series. This remainder is

after n terms of the quotient have been obtained ; and if

l-xx is numerically less than 1, the remainder becomes smaller and

smaller as n increases, while if x is numerically greater than 1,

the remainder becomes larger. Hence in the former case it maybe neglected, but not in the latter case.

The calculus offers a general method for finding such series

as those obtained by the special methods which led to (2) and

(3). This method will be given in the following section.

59. Maclatirin's series. We shall assume that a function can

usually be expressed by a power series which is valid for ap-

propriate values of x, and that the derivative of the function maybe found by differentiating the series term by term. The proof

of these assumptions lies outside the scope of this book. Let us

proceed to find the expansion of sin x into a series. We begin by

writing $oLv s*A+Bx + Cx*+Dx*+Ex*+Fx*+ - --, (1)

where -4, J?, (7, etc. are coefficients to be determined.

174 SERIES

By differentiating (1) successively, we have

- 5 . 4

cosa;=3.2. J> + 4.3 . 2.Je + 5.

sin z = 4- 3- 2- J+5. 4- 3 . 2 . Jfc

-3 -

By substituting x = in equation (1) and each of the fol-

lowing equations, we get

.4 = 0, 5=1, C=0, 3.2.D = -1, J?=0, 5.4.3.2.^= 1;

whence JL = 0, 5 = 1, tf= 0, -O = -^T #=<>, ^=^-o I o

Suhstitutuig these values in (1), we have

8

,X6

SQ^sma; = a;__ + __..., (2)

and the law of the following terms is evident.

The above method may obviously be used for any function

which may be expanded into a series. We may also obtain a

general formula by repeating the above operations for a general

I auction /(a;).

We place f(x) =A+ Bz + C3?+J)x*+lSzl+ - (3)

and, by differentiation, obtain in succession

f"(x) = 3 ! Z> + 4 3 2 . Ex+ ..,

where ff

(x), /"()/'"(); and/v

(a;) represent the first, second,

third, and fourth derivatives of /(#).

We now place a; = in these equations, indicating the results

of that substitution on the left of the equations by the symbols

MACLAUKIN'S SERIES 175

> f" ()> etc - We thus determine A, B, C, D, E, etc.,

and, substituting in (8), have

A*)=AOHAO>+^AoX+^/"(OX+^<oy+-- - (4)

This is called Madaurin's scries.

Ex. 1. Find the value of sin 10 to foui decimal places.

Wo tuny use sones (8), but have? to remembei ( 42) that x must be in

eiivuhii measure. Hence wo i>lace x = -~~ 17458, where wo take fiveloO

significant figmes in ordei to insure accuracy in the fomth significant

iigiU'O of the result.*

Subntibutmg in (2), we have

. TT ,_,_ f.17458)8

,

sin = .17451}- v ' + -ia O

= .17458 - 00089 = .17804.

Ilonco to four decimal places am 10= .1736.

We have u.scd only two terms of the series, since a rough calculation,

which may bo made with a = 2, shows that the third term of the series

will not affect the fourth decimal place.

Ex. 2. Find the value of sin 61 to four decimal places.

In radians the angle 01 is ~-.TT= 1.0647. If this nuxnbei were sub-luO

Htituted in the sonos (2), a great many terms would have to be taken to

include all which affect the first four decimal places. We shall therefore

find a series for8in(~+oA

and afterwards place o: =j~(=l).

We

chow the angle ~^(~ (10) because it is an angle near 61 for which we "

know the nine and cosine. The series may b(5 obtained by the method by

which (2) was obtained, For variety we shall use the general formula (4).

Wo have then . \ i _.

/(O) . tnn|=|V;

This JH not a general rule. In other cases the student may need to canytwo or men three more significant figures in the calculation than are needed

in the result.

176 SERIES

Therefore, substituting in (4), we have

In this we place x = -^- = .01745 and perform the arithmetical calcula-loO . .

tion. We have sin61= sin(^ + ~}= 8746.\o 180/

Ex. 3. Expand In (1 + a)

The function In x is an example of a function which cannot be expandedinto a Maclaurm's series, since if we place /(a;)

= In a;, we find /(0),/'(0),

etc to be infinite, and the series (4) cannot be written We can, however,

expand ln(l + a;) by series (4) or by using the method employed in obtain-

ing (2). The latter method is more instructive because of an interesting

abbreviation of the work We place

ln(l + a;)= A + Bx + Cxz + Dx3 + Eat + .

Then, by differentiating,

= B + 2 CJB + 3 Dx* + 4 Exs + - .

1 + x

But we know, by elementary algebra, that

-- =1 x + xz xs +1 + x

Hence, by comparing the last two series, we have

JB=1, C =-k Z>=, E=-%, etc.

By placing x = in the first series, we find In 1 = A, whence .4 = 0. Wehave, therefore, ^ 8 ^

In (1 + a;)= x -_ + _-_+....

EXERCISES

Expand each, of the following functions into a Maclaurin's series :

1. d*. 2. cossc. 3. tan a;. 4. sin" 1^.

5. tan" 1^ 6. sinf-j + a?).7. ln(2 + aj).

8. Prove the binomial theorem

t . ... i , n(n 1) . n(n 'IVw 2) , .

(a + o;)n=a'l+wan- 1

aj4-v / an~2

a52+ v

Oy

.

v / an-8

a5*4--- .

i\ O\

9. Compute sin 5 to four decimal places

10. Compute cos 62* to four decimal places.

TAYLOR'S SERIES 177

60. Taylor's series. In the use of Maclaurin's series, as givenin the previous section, it is usually necessary to restrict our-

selves to small values of x. This is for two reasons. In the

first place, the series may not converge for large values of x\

and in the second place, even if it converges, the number of

terms of the series which it is necessary to take to obtain a

required degree of accuracy may be inconveniently large. This

difficulty may be overcome by an ingenious use of Maclaurin's

series as illustrated in Ex. 2 of the previous section. We may,

however, obtain another form of series which may be used whenMaclaurin's series is inconvenient.

Let f(x) be a given function, and let a be a fixed value of

x for which the values of /(#) and its derivatives are known.

Let * be a variable, or general, value of x which does not differ

much from a; that is, let x a be a small number, positive or

negative. We shall then assume that/(a;) can be expanded in

powers of the binomial x a ; that is, we write

f(&=A+(x-a) + C(x-a)*+I)(x-a)8-\ , (1)

and the problem is to determine the coefficients -4, B, C, .

We differentiate equation (1) successively, obtaining

In each of these equations place x = a. We have

f(a)~A, /< =, /"(a) =2,1 (7, etc.;

f" (ch f" (a\whence 4=/(a), B =/'(), C^^f

2' #=

8 ,

' et<>-

Substituting in equation (1), we have as the final result

a)^ - - .. (2)

This is known as Taylor's series. Since, as has been said, it

is valid for values of x which make x a a small quantity, the

SERIES

function /(a?) is said to be expanded in the neighborhood of

x a. It is to be noticed that Taylor's series reduces to

Maclaurin's series when a = 0. Maclaunn's series is therefore

an expansion in the neighborhood of x 0.

Ex. Expand In a; in the neighborhood of x = 3.

Here we have to place a = 3 in the general formula. The calculation of

the coefficients is as follows :

/():= In a, /(3) = ln3,

and therefore

1 f~ QNflBT (

- 3)

This enables us to calculate the natural logarithm of a number near 3,

provided we know the logarithm of 3. For example, let us have givenIn 3 = 1 0986 and desire In 3. Then x - 3 =

, and the series gives

= 1.0986 + 1667 - .0139 + 0015 - .0002 + . . .

= 1 2527.

The last figure cannot be depended upon, since we have used onlyfour decimal places in the calculation.

EXERCISES

Expand each of the following functions into a Taylor's series,

using the value of a given in each case :

1. In*, a = 5. M TT4. cos

a;,a -

2. j a = 2.

. TT3. sin x, a =

8. Compute sin 46 to four decimal places by Taylor's series.

9. Compute cos 32 to four decimal places by Taylor's series.

10. Compute e1-1 to four decimal places by Taylor's series.

GENERAL EXEECISES 179

GENERAL EXERCISES

Expand each of the following functions into series in powers of a;

1. ln(l a?). ,1+as . fir'4. ln= 6. sm(v +

2. seco;. 1 -a? \b

3. ,~ 6. cos(-7r + ). 7. /z 3'

8. Verify the expansion of tan a?(Ex 3, 59) by dividing the

series for sin x by that for cos x.

9. Verify the expansion of sec x (Ex. 2) by dividing 1 by the

series for cos x.

1 x10. Expand 1 by Maclaurin's series and verify by dividing

JL ~T~ tC

the numerator by the denominator

11. Expand e*cos x into a Maclaurin's series, and verify by

multiplying the series for & by that for cosaj.

12. Expand e^sime into a Maclaurin's series, and verify by

multiplying the series for tf* by that for sin a;.

13. Expand eTln(l+aj) into a Maclaurin's series, and verify by

multiplying the series for e00

by that for ln(l-|- K).

14. Compute cos 15 to four decimal places

15. Compute sin 31 to four decimal places

16. Compute e* to four decimal places by the series found in

Ex. 1, 59.

17. Using the series for ln(l+ x~), compute Inf to five decimal

places.

18. Using the series found in Ex. 4, compute In 2 to five decimal

places, and thence, by aid of the result of Ex. 17, find In 3 to four

decimal places.

19. Using the series found in Ex. 4, compute In | to five decimal

places, and thonce, by aid of the first result of Ex. 18, find In 6 to

four decimal places.

20. Using the series found in Ex. 4, compute ln$ to four decimal

places, and thence, by aid of the result of Ex. 18, find In 7 to three

decimal places.

21. Compute the value of TT to four decimal places, from the ex-

1 IT

pansion of sin-1a5 (Ex. 4, 59) and the relation sin-1

^=

-g

180 SEEIES

22. Compute the value of IT to four decimal places, from the ex-

pansion of tan-1;*; (Ex. 5, 59) and the relation tan-1- + 2 tan" 1- =

23. Compute -v/17 to four decimal places by the binomial theorem

(Ex. 8, 59), placing a = 16, x = 1.

24. Compute "\/26 to four decimal places by the binomial theorem

(Ex. 8, 59), placing a = 27, x =- 1.

/"* X o-\r\ /*

25. Obtain the integral / dx in the form of a series

pansion.J x

26. Obtain the integral I e'^dx in the form of a series

pansion.^

f*x C&C27. Obtain the integral / in the form of a series expansion.

Uo * ~T" *

28. Obtain the integral /5in the form of a series expansion.

i/o x

expansion

CHAPTER VIII

PARTIAL DIFFERENTIATION

61. Partial differentiation. A quantity is a function of two

variables x and y when the values of x and y determine the

quantity. Such a function is represented by the symbol /(#, #).

For example, the volume V of a right circular cylinder is a

function of its radius r and its altitude h, and in this case

Similarly, we may have a function of three or more variables

represented by the symbols /(a;, #, z),f(x, y, 2, w), etc.

Consider now /(*, jr), where x and y are independent varia-

bles so that the value of x depends in no way upon the value

of y nor does the value of y depend upon that of x. Then we

may change x without changing y, and the change in x causes

a change in /. The limit of the ratio of these changes is the

derivative of / with respect to x when y is constant, and may/ JTJ?\

be represented by the symbol ( ) v^\dxJv

Similarly, the derivative of / with respect to y when x is

constant, is represented by the symbol i-j-Y These derivatives

\dy/xare called partial derivatives of / with respect to x and y re-

spectively, The symbol used indicates by the letter outside

the parenthesis the variable held constant in the differentiation.

When no ambiguity can arise as to this variable, the partial de-

ftf flf

rivatives are represented by the symbols and^-,

thus:

dx \dx A*-*!) Aa;

+ A^>Ay

181

182 PARTIAL DIFFERENTIATION

So, in general, if we have a function of any number of variables

f(x, y, ., 2), we may have a partial derivative with respect to

each of the variables. These derivatives are expressed by the sym-QJ? rvj* A /

' ' ' ' *' orsometimesby/*C*> y>- - *

To compute these derivatives we have to apply the formulas

for the derivative of a function of one variable, regarding as

constant all the variables except the one with respect to which

we differentiate.

Ex. 1. Consider a perfect gas obeying the law v = We may changeP

the temperature while keeping the pressure unchanged. If A* and AJ> are

corresponding increments of t and v, then

P P P, 8v c

and = - .

dt p

Or we may change the pressure while keeping the temperature un-

changed If Ap and Ai? are corresponding increments of p and v, then

A

and *L =-.dp p*

Ex. 2. /= a;8 - 3 x*y + y, Ex. 3. /= sin

(re2 + y

2),

?L = 3 a;2 - 6 xy,

%- = 2 a; cos (x* + w2),-- v ^ y

Ex. 4. In differentiating in this way care must be taken to have the

functions expressed in terms of the independent variables. Let

x r cos 9t y = r sin 6

Then,

or or

fa A fy A- = r sin 6,~ = r cos 6,

c9 dd

where r and 6 are the independent variables.

PARTIAL DERIVATIVES 183

Moreover, since r

8r

Sx

where x and y are the independent variables.

fix

= sin0,

(2)

It is to be emphasized that in (1) is not the reciprocal of in (2).ftr

j /ON r i

and,in(2),=

(

T . j /-INIn fact, in (1),

=^ '

dr

and because the variable held constant is differ-

ent in the two cases, there is no reason that

one should be the leciprocal of the other. It

happens in this case that the two are eqiial, but

this is not a general rule. Graphically (Fig. 80),

if OP = r is increased by PQ = Ar, while 6 is

constant, then PR = Aa: is determined Then

9j, fdx\ PR=( )

\drjaSr

T nLim. - - = cos 0.

PQ,Moreover (Fig. 81), if OM = x is increased by

MN-PQ,= Aa;, while y is constant, then RQ= Ar

,.

-, m,is determined Then

dr /dr\ T .=( )

= Lim8x \dx/v

ST f^y fix

cos0. It happens here that = But -5?- or ox co

in (1), and , in (2), are neither equal nor

reciprocal.

EXERCISESn

Jx

3-1 ,t J35 ZS.M

and ~ in each of the following cases

6 . , =C y

2ajv6. sm -Ji

9, If * = ln<X - 2xy + 2/

a + 3aj -K 2

10. If = Va;2 + 2/

2e-r, prove a; + y

prove ^*.

=

184 PAETIAL DIFFERENTIATION

62. Higher partial derivatives. The partial derivatives of

/(#, y) are themselves functions of x and y which may have

partial derivatives, called the second partial derivatives of /(a-, y~).

rrn, d /3A /2A 2 /9A 2 /9A T> 4. -4. viThey are r ( )' ( TT- ) TT-( )' ^ut ^ mav b shown

17

dx\dxj 8y\dx/ 8x\dy/ dy\dy/J

that the order of differentiation with respect to x and y is imma-

terial when the functions and their derivative fulfill the ordinary

conditions as to continuity, so that the second partial derivatives

are three in number, expressed by the symbols

a/8/\ 3/3A a2/dx\Zy) dy\dx] dxcy

Similarly, the third partial derivatives of /(#, /) are four in

number; namely,

\ j3

/gA"" "

So, in general, T; signifies the result of difEerentiating

y) p times with respect toa?, and g times with respect to

, the order of differentiating being immaterial.

In like manner,p 9r signifies the result of differentiating

/(#, y, g) jp times with respect to #, q times with respect to y,and r times with respect to 2, in any order.

TOTAL DIFFERENTIAL 185

EXERCISES

1. If = (xz + 2/

2

) tan-1^ findx

&2. If e"sm(x y), find

^-3

3V3. If =

Verify ^-^] = -^-[^- m each of the following cases :17

&e\<ty/ By\faJi x

4. * = a:?/2 + 2 ye-

1. 6. s = sm- 1 -.

y

K g-y + y 7 - *

O. w I is> ,

te y Vflj24- j/

2

,82 &*

8. If^ = tan-

9. If In(a;

2 a2

/), prove a2

ji g-g= 0.

O2 / -r/\

10. If V = i cos w<, prove n*r ^^^ + w (w + ^)ii = ^-

63. Total differential of a function of two variables. In 20

the differential of a function of a single variable, y =/(), is

defined by the equation ay=f'(x)dx, (1)

where /* (#) is the derivative of y.

But /(aO=Lim; (2)

and hence, according to the definition of a limit ( 1),

|2=/0)+ e, (8)

where e denotes the difference between the variable ^ and its

limitf (x) and approaches zero as a limit as A > 0.

Multiplying (3) by As, we have

Ay=/(a;)Aa? + Ac. (4)

But Az = dx and Ay =/( + Aas) -/(), so that (4) may be

written in the form

-/(SB) =/() * + e <n (5)

186 PARTIAL DIFFERENTIATION

In the case of a function of two variables, /(, y), if x alone

is changed, we have, by (5),

f(x + Az, y) -/(a, y) = ^dfc+e^ (6)

the theory being the same as in the case of a function of oneOJJ

variable, since y is held constant. The term dx may be denoted

by the symbol dxf.

Similarly, if x is held constant and y alone is changed, we have

/(a, y + Ay) -/(*, y) = ^ dy 4- e2cZy, (7)

dfand dy may be denoted by the symbol d f.

dy

Finally, let x and y both change. Then.

A/=/CB+AZ, y+Ay)-/(a;, y)

==/(+ Aas, y+Ay)-/(z+ As, y)4-/(z+ Az,y)-/(,y> (8)

Then, by (6),

/(a + As, y) -/O, y) = |<fc + ^db; (9)

and similarly, by (7),

/(as + As, y + Ay) -/<> + As, y) = ^dy + e/rfy, (10)OJ? *

where is to be computed for the value(a? + Aa;, y). But if

of yis a continuous function, as we shall assume it is, its value

oyfor (x+ A#, y) differs from its value for (x, y) by an amount

which approaches zero as dx approaches zero. Hence we maywrite, from (8), (9), and (10),

A/= dx + y + ^rfg+ e3rfy, (11)

cj? ay?

where both ^ and ^- are computed for (x, y\dx dy

^ aj

We now write df= j dx + % dy, (1 2)ox dy

so that A/= 4f+ e^+ etdy, (13)

and ?f is called the total differential of the function, the expres-sions dfc/ and ^ being called the partial differentials.

TOTAL DIFFERENTIAL 187

It is evident, by analogy with the case of a function of a

single variable, that a partial differential expresses approximatelythe change in the function caused by a change in one of the

independent variables, and that the total differential expresses

approximately the change in the function caused by changes in

both the independent variables. It is evident from the defini-

tion that , ,. , ,, , ,. ._, A ,.

vf. (14)

Ex. The period of a simple pendulum with small oscillations is

whence g

Let I = 100 cm with a possible error of ^ mm in measuring, andT = 2 sec. with a possible error of ^ $ y sec. in measuring. Then <ll = -^

Moreover, dg = - dl =j- dT,

and we obtain the largest possible error in g by taking dl and dT of oppo-site signs, say dl = ^, dT = -j^.

Then dg =^ + *ra = 1.05 v* as 10.36.

The ratio of error is

<!l = ^ _ 2~ = .0005 + .01 - .0105 = 1.05%.g I T

EXERCISES

1. Calculate the numerical difference between A and dz wliens=s 4 icy a8

ya, a = 2, ?/

= 3, Aas = cZaj s= 01, aud Ay = dy =.001

2. An angle < is determined from the formula < = tan"1^ Toyx

measuring the sides x and y of a right triangle. If x and, y are

found to be 6 ft. and 8 ft. respectively, with a possible error of one

tenth of an inch, in measuring each, iind approximately the greatest

possible error in <.

3. If C is the strength of an electric current due to an electro-

motive force E along a circuit of resistance R, by Ohm's law

188 PARTIAL DIFFERENTIATION

If errors of 1 per cent are made in measuring E and R, find

approximately the greatest possible percentage of error in com-

puting C4. If F denotes the focal length of a combination of two lenses

in contact, their thickness being neglected, and /xand /2

denote the

respective focal lengths of the lenses, then

I= l, i

* A /.'

If/x and/2 are said to be 6 in and 10 in respectively, find approx-

imately the greatest possible error in the computation of F from the

above formula if errors of .01 in. in /tand 0.1 in. infz are made

5. The eccentricity e of an ellipse of axes 2 a and 2b (a > I) is

given by the formula .

The axes of an ellipse are said to be 10 ft. and 6 ft. respectively.

Find approximately the greatest possible error in the determination

of e if there are possible eriors of .1 ft. m a and .01 ft. in 5.

6. The hypotenuse and one side of a right triangle are respectively13 in and 5 in. If the hypotenuse is increased by .01 in., and the

given side is decreased by 01 in,find approximately the change in

the other side, the triangle being kept a right triangle

7. The horizontal range R of a bullet having an initial velocity of

v,fired at an elevation a, is given by the formula

Ji =9

Find approximately the greatest possible error in the computationof R if v = 10,000 ft per second with a possible error of 10 ft persecond, and a 60 with a possible error of 1' (take g = 32).

8. The density D of a body is determined by the formula

*,w w'

where w is the weight of the body in air and w 1 the weight in water.If w = 244,000 gr. and w'= 220,400 gr., find approximately the

largest possible error in D caused by an error of 5 gr. in w and anerror of 10 gr. in w',

HATE OF CHANGE 189QJ7

64. Rate of change. The partial derivative -gives the rate

ox

of change of / with respect to x when x alone varies, and the

partial derivative-j- gives the rate of change of / with respect

to y when y alone varies. It is sometimes desirable to find the

rate of change of/ with respect to some other variable, t. Ob-

viously, if this rate is to have any meaning, x and y must be

functions of t, thus making/ also a function of t. Now, by 11,

the rate of change of / with respect to t is the derivative .

at

To obtain this derivative we have simply to divide df, as given

by (12), 63, by dt, obtaining m this way

dt 8x dt By dt

The same result may be obtained by dividing A/, as given by

(11), 63, by At and taking the limit as At approaches zero as

a limit.

Ex. 1 If the radius of a right circular cylinder is increasing at the rate

of 2 in per second, and the altitude is increasing at the late of 3 in per

second, how fast is the volume increasing when the altitude is 15 in. and the

radius 5 in ?

Let V be the volume, r the radius, and h the altitude. Then

_ ,

y W '

dt~

dr dt dh dt

dr . nd7i+ 7ira -~.

dt dt

By hypothesis,~ = 2,

~ = 3, = 5, 7t = 15. Therefore~= 375 IT cu. in.

T dt dt dt

per second.

The same result may be obtained without partial differentiation by ex

pressing V directly in terms of t. Foi, by hypothesis, r = 5 + 2 1, 7i =15 + 3 1

if we choose t = when r = 5 and 7i = 15. Therefore

7= (8 75 + 375 t + 120 ia+12 *

8)7r;

whence ~ = (375 + 240 1 + 36 2) *-.

When t 0,*- = 375 TT cu. in. per second, as before.dt

190 PARTIAL DIFFERENTIATION

Ex. 2. The temperature of a point in a plane is given by the formula

1

The rate of change of the temperature in a direction parallel to OX is,

accordingly, &*__ 2*to

~"~(a:

2 + jr2)2

'

which gives the limit of the change in the temperature compared with a

change in x when x alone varies

Similarly, the rate of change of u in a direction parallel to OF is

Jw _ 2 ?/

Suppose now we wish to find the rate of change of the tempei ature in a

direction which makes an angle a with OX From Pig. 82, if P^(xv y^) is

a fixed point, and P (x, y) a moving point _on the line through Px making an angle awith OX, and s is the distanceP^P,we have p

P!R = x x1

jo

whence x = xt + s cos a,

"

y = y^ + s sin a,^.

, dx dyand - = cos a, ^ = sma. PIGI 82

Replacing t by s in formula (1), and substituting the values of -- anddii ds~ which we lust found, we haveas

du du du= cos + sin aas dx dy

__ 2 x cos ex + 2 y sin ex

Formula (1) has been written on the hypothesis that x and yare functions of t only. If x and y are functions of two vari-

ables, t and s, and (1) is derived on the assumption that t

alone varies, we have simply to use the notation of 61 to writeat once

a

which may also be written as

dt dx dt %y dt

GENERAL EXERCISES 191

EXERCISES

1. If 2 = etan

x}x = sin

tf, y = cos t, find the rate of change of

with respect to t

1 -f- x2. If = tan" 1

-:~

) x = sin t, y cos t, find the rate of change

of with respect to t when tjL

3. If V= (a*

6""*) cos ay, prove that V and its deiivatives m

any direction are all equal to zero at the point ( 0, )

4. If F= ) find the rate of change of V at the pointVa;2+ y

z

(1, 1) in a direction making an angle of 45 with OX.

5. If the electric potential F-at a,ny point of a plane is given bythe formula V= In Vcca+ y

2,find the rate of change of potential at

any point (1) in a direction toward the origin ; (2) in a direction at

right angles to the direction toward the origin

B. If the electric potential F at any point of the plane is given

-\/(g> _ ff*)2+ W2

by the formula F= In ,

*? find the rate of change of

V(o;H-a)2-|-y

2 b

potential at the point (0, a) in the direction of the axis of y, and at

the point (a, a) in the direction toward the point ( a, 0).

GENERAL EXERCISES

, TJ! xy 1 80 fa A1. If = sin

,

. > prove x -z-- y 5- = 0.

xy + 1 i ox ^oy

"i /?/ /i

2 ' If = S

3. If at = f + ye* prove oj2

-f yr\2 O2

4. If # = e-y cos a(7c cc), prove that

jr-j 4-^ = 0.

B. If 3 = e-c-wx sin /CCj prove that ft= <*.- 1

6. If * = e-**sin (TO?/ + x Vft2 ,2- 7c

2). prove that

v ' x *

2P

aW'rtrr fc /IN *T. * ^^ ,

! Sfr,1 A

7. If F= <?**cos (a In r), prove that -5-5- 4--~ H- 1 -5-75

== 0.N ' or r 0>' ir 0*

192 PAETIAL DIFFEKENTIATION

8. A right circular cylinder has an altitude 8 ft. and a radius

6 ft. Find approximately the change in the volume caused by de-

creasing the altitude by .1 ft. and the radius by .01 ft.

9. The velocity v, with which vibrations travel along a flexible

string, is given by the formula

where t is the tension of the string and m the mass of a unit length.of it Find approximately the greatest possible error in the compu-tation of v if t is found to be 6,000,000 dynes and m to bo .005 gr.

per centimeter, the measurement of t being subject to a possibleerror of 1000 dynes and that of m to a possible error of .0005 gr.

10. The base AB of a triangle is 12 in. long, the side AC is 10 in.,

and the angle A is 60. Calculate the change in the area caused

by increasing A C by 01 in and the angle A by 1. Calculate also

the differential of area corresponding to the same increments.

11. The distance between two points A and B on opposite sidesof a pond is determined by taking a third point C and measuringAC = 90 ft

,BC = 110 ft

,and BCA = 60. Find approximately the

greatest possible error in the computed length of AB caused bypossible errors of 4 in. in the measurement of both AC and BC.

12. The distance of an inaccessible object A from a point B is

found by measuring a base line BC 100 ft,the angle CBA =s<x= 45,

and the angle BCA = 0= 60. Find the greatest possible error inthe computed length of AB caused by errors of 1' in measuring botha and

13. The equal sides of an isosceles triangle are increasing at theuniform rate of .01 in. per second, and the vertical angle is increas-

ing at the uniform rate of .01 radians per second. How fast is thearea of the triangle increasing when the equal sides are each 2 ft.

long and the angle at the vertex is 45 ?

14. Prove that the rate of change of * = In(a; -J-Vjc

a+ f) in thedirection of the line drawn from the origin of coordinates to anypoint P(x, y) is equal to the reciprocal of the length of OP.

15. The altitude of a right circular cone increases at the uniformrate of .1 in per second, and its radius increases at the uniform rateof .01 in. per second How fast is the lateral surface of the coneincreasing when its altitude is 2 ft. and its radius 1 ft.?

GENERAL EXERCISES 193

*t-

ffi ~\ JL* >

16. Given = tan" 1

;

1- tan"1 Find the general expres-

sion for the derivative of along the line drawn from the origin of

coordinates to any point. Find also the value of this derivative at

the point (1, 1).

17. In what direction from the point (3, 4) is the rate of changeof the function tt kzy a maximum, and what is the value of that

maximum rate ?

18. Find a general expression for the rate of change of the func-

tion u s= &-v sin x + - e-*v sin 3 cc at the point ( -77 ) ). Find also the3 \3 /

maximum value of the rate of change.

CHAPTER IX

INTEGRATION

65. Introduction. In 18 and 23 the process of integration

was defined as the determination of a function when its deriva-

tive or its differential is known. We denoted the process of

integration by the symbol /; that is, if

then Cf(x) dx = F(x) + C,

where C is the constant of integration (18).The expression f(x) dx is said to be under the sign of inte-

gration, and/(a;) is called the integrand. The expression J<\J'}+ ('

is called the indefinite integral to distinguish it from the definite

integral defined in 23.

Since integration appears as the converse of differentiation,

it is evident that some formulas of integration may bo found

by direct reversal of the corresponding formulas of differentia-

tion, possibly with some modifications, and that the correctness

of any formula may be verified by differentiation.

In all the formulas which will be derived, the constant C will

be omitted, since it is independent of the form of the integrand;but it must be added in all the indefinite integrals found bymeans of the formulas. However, if the indefinite integral is

found in the course of the evaluation of a definite integral, the

constant may be omitted, as it will simply cancel out if it has

previously been written in ( 23).The two formulas / *

I cdu = c I du (1)

ft

andi >,

- -.^

/

194

](du+ dv + dw-\ ) = Cdu + Cdv + Cdw + . (2)

INTEGEAL OF w* 195

are of fundamental importance. Stated in words they are as

follows :

(1) A constant factor may be changed from one side of the sign

of integration to the other.

(2) The integral of tJie sum of a finite number of functions is

the sum of the integrals of the separate functions.

To prove (1), we note that since cdu d(cu), it follows that

I cdu = I d (cu) cu = c I du.

In like manner, to prove (2), since

du -H dv + dw + . . . = d(u + v + w +...),we have

/ (du + dv + dw H- )= I d (u + v + + )

u + v + w+I du -H

Idv + I dw + . . .

The application of these formulas is illustrated in the follow-

ing articles.

66. Integral of u". Since for all values of m except m =

/wm\or d )=wm

~

\m/

it follows that Cum~ l du = .

J mPlacing m == n +1, we have

/or aZZ values of n except n s= 1.

In the case %= 1, the expression under the sign of inte-

gration in (1) becomes , which is recognized as

Therefore T = lnw. (2)

196 INTEGRATION

In applying these formulas the problem is to choose for u

some function of x which will bring the given integral, if pos-

sible, under one of the formulas. The form of the integrand

suggests the function of x which should be chosen for u.

Ex. 1. Find the value ofJYaa;2 + Ix + - +

J

dx

Applying (2), 65, and then (1), 05, we have

fftJ \

= afxz(lx +

bj'xflx+ cf

(~- + kJVThe fiist, the second, and the fourth of these integrals may be evaluak'tl

by formula (1) and the thud by foirnula (2), wheie u = x, the icsults beiny11 &respectively

- ax9,-

bx*,--

, and c In a..-,

o &

Therefore C lax* + lx + - + -^} dx = ~ ax* + J lxz + c In i - - + CJ \ x a,*/ 3 2 x

Ex. 2. Find the value ofJ*(z

2 + 2)ar<fo.

If the factors of the integrand are multiplied together, we have

f(a;

2 + 2) xdx = f(xs + 2

a;) dx,

which may be evaluated by the same method as that used in Ex 1, the

result being z4 + xz + C.

Or we may let x2 + 2 = , whence 2xdx = du, so that xdx = % du. Hence

22

Comparing the two values of the integral found by the two methods of

integration, we see that they differ only by the constant unity, which maybe made a part of the constant of integration.

Ex. 3. Find the value of C(oxz + 2 bz)

a(ox + &) dx.

Let GKC2 + 2 bx = u Then (2 ax + 2 fydx = du, so that (ox + &) dx = J rfw.

HenceJ" (oa;

2 + 2 fcc)8(aa: + l)dx = C %u*du

1 /7 1 u4

,=s / MBdu = . + C2J 24

INTEGRAL OF M" 197

Ex. 4. Find the value of C4 < +

*]

dx

J ax2 + 2 fa;

As in Ex. 3, let axz + 2bx = u. Then (2 ax + 2 b) dx =* du, so that

Ilence

= 2 In w + Cs= 2 In (az

2 + 2 to) +

Ex. 5. Find the value of C (e +

Let e * + & = u. Then eaxadx = tfu.

HenceJ(e

+ 6)cdiB J^ jf

If the integrand is a trigonometric expression it is often pos-

sible to carry out the integration by either formula (1) or (2).

This may happen when the integrand can be expressed m terms

of one of the elementary trigonometric functions, the whole

expression being multiplied by the diffeiential of that function.

For instance, the expression to be integrated may consist of a

function of smo; multiplied by cosEefe, or a function of cos a;

multiplied by ( sma^a;), etc.

Ex. 6. Find the value of [Vsm x coasxdx.

Since d(smx) = oosxrfx, we will separate out the factor cosrcrfar and

express the rest of the integrand in terms of sin a:.

Thus -Vsinxcossxdx = Vsirue (1 sms;c) (cosxdx).

Now place sin x = M, and we have

198 INTEGRATION

Ex. 7. Find the value ofJsec

62xdz.

Since d(tan 2a:)= 2 sec22 xdx, we separate out the factor sec82 rdc and

expiess the rest of the integrand in terra a of tan 2 a:

Thus sec 2 xdx = sec*2 x (sec32

=(1 + tan82

a:)2(sec

3 2 a:r/

= (1 + 2 tan22 ar +*tan*2 a;) (se

Now place tan 2z=su, and we have

fsec62 xdx =

= Jtan2a:

EXERCISES

Find the values of the following integrals

*6M 4- sec3 ace

tan ax

+ C.

. C

-j=\dx.

C( r 1 \^J [x-vx j=]dx.J \ x~vx/

r?**-C tffa

"J^Ti. C(x*+ l)*xdx.

I Vo4 + 4x8t7a;.

/.cJ 2 a; -f sin 2 a;'

in f 1-cosa;!. I . . .,dx.J (aj sinaj)

4

-/:12.

13

dx.

&r.

sin axrfa;.

e8*^

e*x + &

_,

1 + COS 2 G ,9. I - - - ote.

fifJ 1 + cos aa

14. I cos82 as sin 2 OJ$B.

16. I sin33 a; cos 3 xdx.

16. / sin (x + 2) cos (a -|- 2) Ja;.

17. / cos^3 x sin 3 *c?a;.

18. I sec4 3ic^a5.J

19. rctn2(2a;-|-l)cso!1

(2aj

"*/'

ALG-EBKAIC INTEGRANDS 199

67. Other algebraic integrands. From the formulas for the

differentiation of sm~ 1M, tan" 1

^, and sec-'w, we derive, by re-

versal, the corresponding formulas of integration :

du== sin" 1

?/,

duand

These formulas are much more serviceable, however, if u is

/

replaced by -(a > 0). Making this substitution and evident

fit

reductions, we have as oar required formulas

du . ,u .^ N== sin" 1 -(1)

<J <J /y NXa u aL

_, -ON= -tan- 1 -> (2)2 s y

and

+ a2 a a

du

/du1 *u sQ ~

^-sec" 1 -- (3)wVwa- a* a-

Referring to 1, 47, we see that sin" 1 - must be taken in theUf

first or the fourth quadrant; if, however, it is necessary to

uhave sin"

1 - in the second or the third quadrant, the minus sign

must be prefixed. In like manner, in (3), sec"1 - must be taken in

the first or the third quadrant or else its sign must be changed.

/fix__ Letting 2

Vj) 4 a:2

r fa _. f \&uJ

-y/9 _ 358J V9 ~ M3

/fix__ Letting 2 x = u, we have du = 2 dx;

whence dx = ^ du, and Vj) 4 a:2

200 INTEGRATION

dxEx.2 Find the value of f

,

*Jf we let V:JW~ M, UH-J

Jjt V;} j,

a - 4t

du = V3 rfj; ;whence <ir = -= f/, and

Va

/^j _ r__ <lu

x V3 x2 -4 ^ w vV~-~4

/ rfr

Ex. 3. Find the value of I /.17

Since V4 a; jsz=Vd (r 2)

2, we may lot w =s a 2

;whttiusH ^/.r

-*/, and

C ^ C_~ t

J - 2 "^ ^*

S1J1,

-i + C'6

x -~ 2

Ex. 4. Find the value of C<lx

J 2 .r + 3 u, + 5

We may first write the integrand in the form

1 1 = 1 1

2 aJ+x+ 2'*and let u = x + I Then du = dj.,

J 2 2 + 3 x + 5=

<>J(.r + ^)

B+"Y

= ~f2J i

1 Xs ton-'JL

4 4

_ 2t4tt

-~r~ auv"f5f

+r

-Vai

/

ALGEBBAIC INTEGRANDS 201

5z-2 ,

Separating the integrand into two fractions

5z _ 2

2o;a 4-3 2a;2 +3'

and using (2), 65, we have

2 / 5xdx r 2dx__/

J

If we let rz = 2 a;2 + 3, then du = 4 artfo

5a;c?a; 5 /vM 5, 5

and if we let u V2 x, then rfu = V2 dx

, r 2dx AT / du /= 1 . . it Vo. ,a;Voand I i4= V2 I - = V2 - tan- 1

^ = tan- 1 -

J2a;2+3 Ju*+$ V3 3 3

Ex. 6. Find the value of

There is here a certain ambiguity, since tan-1V and tan~ 1( 1) have

each an infinite number of values. If, however, we remember that the graphof tan~1

a; is composed of an infinite number of distinct parts, or Iranches

(Fig 56, 46), the ambiguity is removed by taking the values of tan- 1Vsand tan- 1

( 1) from the same branch of the graph For if we consider

- 1^ tan- xo and select any value of tan~ 1a, then if & = a,

/

|Jaa X + 1tan~ l6 must be taken equal to tan-1

^, since the value of the integral is

then zero. As & varies from equality with a to its final value, tan-x & will

vaiy from tan-1a to the nearest value of tau-a&.

The simplest way to choose the proper values of tan-1?* and tan~ 1 is

to take them both between and Then we have

/'Va dx _ v__

/ TT\ _ 7

./-i a2 + 1~

3 \ i/~ 1TT

12*

The same ambiguity occurs in the determination of a definite integi al

by (1), but the simplest way to obviate it is to take both values of sin- 1

7T 7Tbetween

-^and The proof is left to the student.

B

202

EXERCISES

Find the values of the following integrals :

r dx r dxl '

J Vl6 - 9 a?'

J V5 x - 3 a/

dx dx

/dxC B

a- 12. /, -

/dxC (?SK

13 I

a:V4a;2 -9'

J 3-e3 - 4 a; + 2*

ra.r + 11

x 1^

16 '

7 '

a

^J

10 '

r /

*19 rJ VGas-a* 19 '

J_

T^

oJ V6*-4a*

68. Closely resembling formulas (1) and (2) of the last sectionm the form of the integrand are the following formulas ;

+o. (1)

/du>-. / /

^r^^(u+^u --^ (2)

and

These formulas can be easily verified by differentiation andthis verification should be made by the student'

ALGEBRAIC INTEGRANDS 203

/f/1".

Letting V2 a; = u, we have du = "v2dx;whence dx = du, and

,;= du

= _JL T rfu

= -i- In [M + Vu*-8] + CVa

= 4= ln C'^2 + Va xa - 3] + a

rfe

/(/~ ~

Vy a B

As in Ex. 4, 07, we may write tho integrand in the form

Va Vxa-f- 3 a? Va V(j; + ^)

a ~$

and lot M = x + S !whence rfw = rfx,

/__.4!L_JL T - ^-

;

V3 a:a + 1 x VJJ ^

V(,^ + *j)a-

J

V8^ Vua^

"vl

n w

"vl~ -Lin (3 + 2 + VO a;

2 + 12 *) +K>Vi)

where C = -- In 8 + A".

Va

/o i IKz a;"8 + * JLO

Writing the integrand in the form

I ^j1

,..^ "Swe let * x Hh 1 ;

whence rf ss rfa?.

204 INTEGRATION

r dx _ 1 f* <lf

J 2s* + a: -15~

2 J (* + i)a W"

2 2(-V) + V

llnljli + C11 a + 3

It I >1'

11 X + i)

where C = -j^ In 2 + K.

EXERCISES

Find the values of the following integrals :

1 r *"11 r~ /* ~j.i

-.

* J.J.. i,

-

J V^T2 J 3^ + 6

2 c dx12 r ^

J V9rf-l' Jo*-3(B +

3 r*"

13 r rfa!'

J V3 2-4'

J *+-* C dx C <fa

'

J Va^a"'

J 4if-2* -

5. f^

IB f*rf

f-'

J V3jca +2a3 + 3'

J V5-""4'

6C dx r* dx

J***-M '

J VOSTT'r dx

'I oa -i' 1^ 2a^- 1

'-a- 8'

TRIGONOMETRIC FUNCTIONS 205

69. Integrals of trigonometric functions. Of the following for-

mulas for the integration of the trigonometric functions, each

of the first six is the direct converse of the corresponding for-

mula of differentiation ( 44), and the last four can readily be

verified by differentiation, which is left to the student.

/ sin udu cos M, (1)

Icos udu = sin w, (2)

I seu"ud (, = tan u, (8)

/Qscfudu= ctn w, (4)v J

sec u tan udu = sec it, (5)

esc u ctn udu = esc u, (6)

Itan udu = hi sec u, (7)

I ctn udu = In sin it, (8}

|sec udu = In (sec u -h tan it), (9^

Iesc udu = In (esc it ctn u) .

Ex. 1. Find the value of Tsui 7a.v7a;.

If we let 11=7,1:,

then du7dr;whence dx = \ du,

and Cs\n7xdx = T sin (| f/u)

=3 ^ Isin udu

=\ cos + C*

=J-cos7a? + C,

206 INTEGRATION

Ex. 2. Find the value of fsec (2 x + 1) tan (2 x + 1) dx

If we let w = 2 a; -f 1, then du = 2 dx,

and Tsec (2 x + 1) tan (2 x + 1) cfo = fsec M tan du

= ^secu + C= sec (2 a; + 1) + (7.

Often a trigonometric transformation of the integrand facili-

tates the carrying out of the integration, as shown in the

following examples:

Ex. 3. Find the value of f coazaxdx

Since cossax - (1 + cos 2 ax),

Ccos*ax dx = f (J + J cos 2aar) dx

= Tdk + ^ Tcos 2 aacfoJ

i/

=o x + T~ sin 2 aa: + C1

,

A T CZ

the second integral being evaluated by formula (2) with M = 2 ax.

Ex. 4. Find the value of fVl + cos arcfo.

Since cos x = 2 cos2- 1,ii

Vl + cosa;=v/

2cos|,- ib

and fVl + cos xdx ~ CV^ cos - </j;

Ex. 5. Find the value of Aan23 xdx,

Since tan23a; seca3 -1,

Jtan23 arcfa: =

J*(sec23 a: - 1) da:

= Csec*3xdx Cdx

=J-tan3a; a;,

bhe first integral being evaluated by formula (3) with v*=

TRIGONOMETRIC FUNCTIONS 207

EXERCISES

Find the values of the following integrals :

22.a;.v'10.

Jose9

(3 2a:)

da

rr,, /*cos2o; , 23. I (

.yll. I : dx. ] ir*J smaj J~i

r . * r*i12. I sina7rdaj. 24.

J

J 2 Jis

70. Integrals of exponential functions. The formulas

/ It J M XI \

and |avdu*=-. a* (2)

J In a

are derived immediately from the corresponding formulas of

differentiation.

208 INTEGRATION

Ex. 1. Find the value of C$ x djc

If we let 3 x = u, we have

Ce8x dx

/*/BJLrrfa;.xz

xi

iIf we place V5 = 5"" and let = u, we have

._/*.

EXERCISES

!Fmd the values of the following integrals :

1. ie^^dx. 6. f(a + -)<&. 9. f 10*dx.

//g2x _ a-2as /

e^ajtfo;. 6. / _ a,rf. 10. / 200

11. C\**-*dx.

. fV~

*"'rfaj.

Jo ^+0"*

8.

4. e^f+txfa Qt +e"dx. 12

71. Substitutions. In all the integrations that have beenmade m the previous sections -we have substituted a new vari-

able M for some function of x, thereby making the given integralidentical with one of the formulas. There are other cases in

which the choice of the new variable u is not so evident, butin which, nevertheless, it is possible to reduce the given integralto one of the known integrals by an appropriate choice and wh>stitution of a new variable. We shall suggest in this section a fewof the more common substitutions which it is desirable to try,

I. Integrand involving powers of a+ U. The substitution ofsome power of 2 for a -f- bx is usually desirable.

SUBSTITUTIONS 209

Ex. 1. Find the Value of f^^

,.

J(1+2*)*

Here we let 1 + 2 x = s?;then x = ^ (

8 -1) and dx = % z*dz.

Therefore Cx* dx = -

f(z<- 2 * + z) tls

17

(1+2 a?)*8J

Eeplacing a by its value (1 + 2 #)* and simplifying, we have

=(1 + 2 B)

*(" 12 * + 2 ^ + c'

II. Integrand involving powers of a + ?;#". The substitution

of some power of for a + frr71

is desirable if the expressionunder the integral sign contains xn

~ l d& as a factor, since

= bnxn~ 1

cfo.

/x/j.2

J. x 2

rfa;.

a;

We may write the integral in the form

and place aa + o2 = a

. Then xdx = sdz, and the integral becomes

Replacing s by its value in terms of x, we have

x 2 Vaa + a2 + a

Ex. 3. Find the value offjs

6(l + 2 spfidx.

We may write the integral in the form

and place 1 + 2 a;8 = 2

. Thou xhlx = ifisdz, and tho now integral in z ia

frJ(*<- *a) ds

= n^ s (3 *2 ~

5) + C.

Replacing z by its value, we have

Cx6(1 + 2 a)icfa ~ ^ (1 + 2 38

)8 (3 a8 -

1) + C.

X

210 INTEGRATION

III. Integrand involving Va2ar

2

. If a right triangle is

constructed with one leg equal to x and

with the hypotenuse equal to a (Fig. 83),

the substitution x = a sin is suggested.

Ex. 4. Find the value of fVa8a:2da;. _, gg

Let a; = a sin <. Then dx = a cos<f> d<f> and, from the triangle, V 2

a;a

= a cos <.

Therefore C^/a?x*dx = a2 fcos2

But

and

for, from the triangle, sin d> = - and cos <& =a a

Finally, by substitution, we have

J 2 \ a/

IV. Integrand involving V^+a3. If a right triangle is

constructed with the two legs equal to xand a respectively (Fig. 84), the substitu-

tion x = a tan<f>

is suggested.

Ex. 5. Find the value of fdx

i/ /* o * ^o\ $

Let i = a tan < Then dx = a sec8 <tf< and, from the triangle, V#a + aa= a sec

<f>.

Therefore f

But, from the triangle, sin<f>= *

;so that, by substitution,

Va;8 + flra

*

SUBSTITUTIONS 211

V. Integrand involving V#a a2. If a right triangle is

constructed with the hypotenuse equal to

x and with one leg equal to a (Fig. 85),the substitution x = a sec

<#>is suggested.

-*f rr a

Ex. 6. Find the value of / a8 Vo:2 a?dx"

^^ ge

Let as = a sec <. Then dx = a sec < tan < d< and, from the triangle,Va;2 - a2 = a tan

Therefore fa8 Vj;2 - a2dx = a

= aG

J"(tan2

</> + tan4<) sec2

.J._

j

Exit, from the triangle, tan <j>=

,so that, by substitution, we have

J>V*2 - aa rf.c = TV (2 aa + 3 a;

2) V(a;

2 - a2)8 + C

We might have written this integral in the form fa;2Va;fl

cPfxdx) andsolved by letting za = a;

2 a2.

72. If the value of the indefinite integral is found by substitu-

X6f(x)dx may be

performed in two ways, differing in the manner in which the

limits are substituted. These two ways are shown in the solutions

of the following example:

Ex. Find f Va"--.ida;.Jo

By Ex. 4, 71,

Va x*dx = i/a Va - x* +

Therefore C"

212 INTEGRATION

Or we may proceed as follows Let x = a sin < When x = 0,=

;

and when x = a, < =, so that < varies from to - as x vanes fiom to a.

Accordingly,"

v

4

The second method is evidently the better method, as it obviates the

necessity of replacing z in the indefinite integral by its value in tenna of

x before the limits of integration can be substituted.

EXERCISES

Find the values of the following integrals :

1. \

X.

*6. I

~~. 11

'/(*-*7. r **<

J (3-

/,\J &

9. f*

J (V5 -

5 -I . .

, 10. I aV2a; $dx. 15

73. Integration "by parts. Another method of importance in

the reduction of a given integral to a known type is that of

integration by parts, the formula for which is derived from the

formula for the differential of a product,

d (uv) = udv + v du.

From this formula we derive

uv = / udv + f vdUy

which is usually written in the form

I udv = uv I vdu.

INTEGEATION BY PAETS 213

In the use of this formula the aim is evidently to make the

original integration depend upon the evaluation of a simplerintegral.

Ex. 1. Find the value of Cxe*dx,

If we let x = u and cPdx = dv, we have du = dx and u = e*.

Substituting in our formula, we have

j x&dx = are1j

exdx

It is evident that in selecting the expression for dv it is desirable, if

possible, to choose an expression that is easily integrated.

Ex. 2. Find the value of Csm

Here we may let sin-1 x = u and dx = do, -whence du =rx

and v = x.

Substituting in our formula, we have "^ 1 ~~ x*

xdxsin-i-xdx = x sin-1 a:

{J Vl-o;2

= x sin-1^ + l a;2 + C,

the last integral being evaluated by (1), 66

Sometimes an integral may be evaluated by successive inte-

gration by parts.

Ex. 3. Find the value of Cx*e*dx.

Here we let xz = u and eFdx = do. Then du ^xdx and = e31.

Therefore C

The integral f xeFdx may be evaluated by integrationby parts (see Ex. 1),

so that finally

Ja;V=dx = x*<*> - 2 (ar

-1) e* +C = eP

(a;2 - 2 x + 2) + (7.

Ex. 4. Find the value ofje"

sin&zda:.

"

sin bx = u and' c** rfz = dv, we liave

=- *<"* sin &K fc8* cos Ba; ?.

a v

214 INTEGRATION

In the integral Ce01 cos bxdx we let cos bx = u and eaxdx dv, and havi/I b /*

e"* cos bxdx - e"* cos fa + -/ e** sin Sards.

a a w

Substituting this value above, we have

/e"*sin bxdx = - e01 sin &e -- (

- eP* cos &r + - fc * sin bx dx\a a\a aJ )

Now bringing to the left-hand member of the equation all the terms

containing the integral, we have

(1 + ) fe * sin bxdx = - e"* sin bx ---e031 cos bx,\ a?/ J a a?

t. r __ , , fi** (a sin bx b cos bx)whence I e"* sin bxdx = v '

J az +b*

Ex. 5. Find the value of fVa;2 + a?dx.

gPlacing Va;3 + c2 = w and <?a; = fo, whence rf =

we have Va;B + oaand w = a,

Since a;2 =

(a^ + a2) a2, the second integral of (1) may be written as

/a;2 + a2

which equals z2 + as/*

Evaluating this last integral and substituting in (1), we hare

whenceJVa;

2 + a?dx =s $ [a Var2 + aa + aa In (x +

74. If the value of the indefinite integral f/O) ** is found

by integration by parts, the value of the definite integralr*I f(z)dv may be found by substituting the limits a, and I inva

the usual manner, in the indefinite integral.

INTEGRATION BY PARTS 215

IT

Ex. Find the value of C*x*sinxdx.Jo

To find the value of the indefinite integral, let a;2 = and sin a; da; = dv.

ThenJ

a;asm xdx = a?

2 cos x + 2Jx coaxdx.

In / a; cos a, dx, lot a; = u and cos xdx = do.

Then \x cos x dx x sin x i sin x dx

= x sin a; + cos x.

Finally, \ve have

I x*ainxdx = ara cosa, + 2 a? sin a: + 2 cos a; + C.

IT

"" P HaHence C*j? sinarda; a;

2 cos x + 2 x sin x + 2 cos a;

= -2.

The better method, however, is as follows: .

l$.f(x)dx is denoted by udv, the definite integral \ f(x)dxstb Jo.

may be denoted by I udv, where it is understood that a and b

t/a

are the values of the independent variable. Then

/& . s*b

I udv=z\uv~\ n Ivdu.

Jo, v a

To prove this, note that ib follows at once from the equation

, /^6 />fc /*! /&

** / / / /i/a yo /o vo

Applying this method to the problem just solved, we have

IT

v r -13 ir

r'5'a;a sina?afa;= a;

2 cosa; + 2 C^xciOBxdx

/* * j

Jo

["|JW

2a;sina: 2 C*&inxdx

w

STT-f 2 cos a: I

L Jo

216 INTEGRATION

EXERCISES

Find the values of tho following intftfra

. CynP'tlr. 5. I .rscc" 1 !}^//^. 9. I ./ *-V.r.

. I stP<i*L(?x- 6. I (In Hindoos ;'(/.r. 10, / ,i"'lu.i'f/>.

. COOB- IX<?JK. 7. I/'9 "Von .r /.i'. il. / Mr "J, !/,!.

|tair~

13a:</J3 8. / ,r uos3

'^<Ar. 18, / VPOH JJuv/.i".4.

75. Integration of rational fractions, A rtttwixtt fwrfiHit i.s it

fraction whose numerator and denominator ur polynomials. Il

can often be integrated by expressing it UH tho twin of

fractions whoso denominators are faot-oi'H of th

of the original fraction. Wo .shall illustrate only tin* inw in

which the degree of the numerator is lews thun tho th'gm* of tin*

denoiuinator and in which the ftu'torH of tho tlmitmunutur m*

all of the first degree and all different,

Ex. Find tho value off-~^ f^}'\"

</''

The factors of tho denominator arc x + tt, x 2, a<l x + 2, Wo ftHwtmo

(x + 8)(j- 4) ""a +a-B^ + a

whojo J, /?, and ^7aro constants to bo (lotonniucd.

Clearing (1) of fmclions by uiuUiplyuig by (ja + n)(x9

4), w

or

l, /J, and (7 firo to bo datidrmmnd so thai Iho righWrnml memhrof (3) .shall bo identical with th k'ft-hand nuwnbcr, lhn <'<*iHj5iHtu( wf

powers of a: on the two sides of tho equation muni Iw tujual.

Therefore, equating the ooeffieimitH of liku powors of / in(Jl),

theequatzons

- 1 A + <t 7i ~ G ~ li,

whence we find ^t =~ 2, Ji 2, r,Y

= 1,

RATIONAL FRACTIONS 217

Substituting these values in (1), we have

a?8 + 11 x +14 _ 2 2 1

= - 2 In (x + 3) + 2 In(a;-

2) + In(a, + 2) + C

EXERCISES

Find the values of the following integrals

i r a;+i7 A r*'

J ?-"o7+8'^ 4 '

J ( 2) (*-Bas+l

T__ 2 -5a! + 5 T'

J (iC -.l)(aJ ~2)(a! -3)

rfa; - 6>

J

76. Table of integrals. The formulas of integration used in

this chapter are sufficient for the solution of most of the prob-lems which occur in practice. To these formulas we have added

a few others. In some cases they represent an integral which

has already been evaluated, and in other cases they are the

result of an integration by parts. In all cases they can be

verified by differentiating both sides of the equation.These collected formulas form a brief table of integrals which

will aid in the solution of the problems in this book. It will be

noticed that some of the formulas express the given integral onlyin terms of a simpler integral.

I. FUNDAMENTAL

1.|cdu=sc

Idu.

2. I (du + Av + dw )= Idu +

|dv + I dw .

8. / udv sauv I vdu.

213 INTEGRATION

II. ALGEBRAIC

4./undu = . (n = 1)

J n + l

C du5. I

= mu.

du 1

p^ = ln

rA I

'

J_ r du _ 1 , u a'

\ ~o 5 o ^!

'

J u a za u-\-a

8. / vfls u du = ( u ~\a?

JV

2V

9. JWa2-wadM =-i.Oa_t

ju^a*-u*du=>- w + 2

sm~-

19.

w

10.

12.

13.I __ / t>.

n n/ . _

14.

15.

16.

/duVa2-

fl2

/*

Ls*

Iu^/uz azdu = % (w

2

TABLE 219

_. du 1 ,M20. I . =-seo- 1 -

/,

.

IV2 au u*du = -

(w a) V2 au w2+ a2sin"

1

/.22. / t

""* = sm- 1 -

V2 aw - w2 a

III. TRIGONOMETRIC

23. I smudu=x cosw.

ft. /* . , , u 1

24. I sm waw == - -r sm 2 w.

J 24/I w, 1 /*smnudu= sin""^ cos w H I sin""

2wcZM. (n 3= 0)

w w J^ J

26.|cos udu SB sin w.

/w 1cos

2wdw == - + T sill 2 w.2 4

/I 7i _ i /

coa*udu BBS - cosn-1w sin w H I cos"

~ aw^. (w ^ 0)n n J

29.Itan udu In sec w.

/tan'*"''^C H /\\

tanww?w s=rf I tanw

""udu. (n l^* 0)n-1 J

81. f ctnw rfw sss In sin w.

/r*i"n" ***

i/ /**

ctnnw^ - r - I ctn11- 3

wc?w. (n- 1 + 0)n-1 J

88.Isec wrfti a= In (sec t* + tan w).

34.Jseo'wtfw

=s tan M,

85,Jcso

WC^M In (cso w - ctn ).

86,Icso

awc?'M! ctaa w.

220 INTM! RATION

37. I sec u tau u du HUC w.

38. I oso ? ctn w tft* = <IBO ?f.

/"

/

/Hinmfl?M'0,s"

1 U H -I r .

HillmWCt)Sn

?/,(Zjts=S" '-'-.-,} / ,sh'"M riLS* "'/0/,m + w / 1 ><</

( i i it a)

I || I U 1

|'ll"*1/ I"*/ VU*' O/ //')/ " - -

j., U4_ i I Wtll // 4*1 i 1!* J/ */Jjf* v I niix c* ^v'Ci (v vvCv "*-"*^ T i Ji i * F**'I ~ *i**

Jsi,n I , j M11141 . I Hill"'W C50S" U dlt,*x

42. Tsii

m + 1

IV. EXPONENTIAL

r<'"?MM".

4. Ca*du**-~aH.

J In a

43.

44.

45.

46.A*

,

IWCOH" H M

< H- 1

{/ f 1 / >

GENERAL EXERCISES

Find the vaJuos of tho following mtgralH t

I.

J (3aj84- 4 -

^3~^ flfaj. 4, f(j f

. f2aja

J

GENERAL EXERCISES 221

7. f(2-

9. fi+ *2

* w- rJ (3x + xrf J

r^-. 27. rJ as - 1 J (//n-(8.-l)orf(8.-l)*, "

J (

IS.

13.

dx

dx

/C dxGSG^4; C l/C 30 I

*

/ "\/ 1\ 'V* r- -

11 '>*i/ v ti a/ "

t*/

11. fsec8(-2Han8

(:-2)daJ. 31. C ,

dx

J V 'J V34-2aj-o;a

15. / ctn(a; 1) sec

4

(a; 1) dx. 32 .|J J W .

1C. f oso82ajctus2a;^sB. 33.

J

17.Jtan'Saj-yseoSa-rZa-. 34 .

T- &

/H -U-Jlu-.iniLJ.UI

(itn 2 IB V(!HO '2xdx. 35.

/-..-.-.IIU.J. rt

CMC* fiaJ"V(!t,u 5 "/?. 3g / _*

I . .<!()

J '

/***"

*?*"" $$, o*y I

siu*4aj"

J 4. s

/CHC

4 5 C /*^ 38t/

v tan8Sir J >!'

M.J-*.v 39

'/9> +1

19

20.

222 INTEGRATION

58.

*

54.

55,

/

53. / --'*,&,70t

/ ^/

/^fal "71 I

V3a:2+ l' /J

"J

/<a;^

V2a;a- 3aj'72 '

J seo(a!- |)<fa.

56. /^dx

_ ^8^43;^/ v4.'ra _L /Lo _l_7 ' d<

I ?T~ <W5>*yv * a; T*a:-f-<

^y cos 2 x

/OnKSI t

. . w . i /sin 2 a/ cos/w a / 4 I I -- _^ j_"V n <T^ _I_ /Lrw'i * ^ * I I ,

***"' ' "voa+4a;-l J \ Sinas COSJB

75. Cam*~dx

GENERAL EXEECISES

91. Cx5*+*dx.

223

89.

106.

224

107.

INTEGEATION

ptan(as-J \ Q

na108.

109.

110.

111

112

Bx ,Bx

114.|

i/S

*.

a;Va;84- 9

115 -

116.

118.

ofo:

JVt/Q

123. I acHarr^cfaj..

1Jo

CHAPTER X

APPLICATIONS

77. Review problems. The methods in Chapter III for de-

termining areas, volumes, and pressures are entirely general,

and with our new for- ymulas of integration we can

now apply these methods

to a still wider range of

cases.

Ex. 1. Find the area of the

x* i/"

ellipse -^ +~ = 1.

It is evident from the sym-

metry of the curve (Fig. 86)that one fourth of the required

aiea is bounded by the axis of pIG< gg

y, the axis of x, and the curve.

Constructing the rectangle MNQ,P as the element of area dA, we have

dA**ydx = --VtP-aPdx.a

Hence

5ST

Ex.2. Find the area bounded by the~

axis of x, the parabola y9 =a kx, and the

straight line y + Sff-frssO (Fig. 87),c

The straight line and the parabola intersect at the point C ( j s )> and

/, \ \4 -i/

the straight line intersects OX at B ( - ,

j. Draw CD perpendicular to 0Z.

If we construct the elements of area as in Ex, 1, they will be of different

226 APPLICATIONS

form according as they aie to the left or to the right of the line CD] fc

on the left of CD we shall have

dAydx k*x*dx,

and on the right of CD we shall have

dA = ydx = (k 2a:)

dx

It will, accordingly, be necessary to compute sepaiately the areas ODi

and DBC and take their sum.

Area

AreaDBC =J*

3(X:- 2 x) dx = \kx - a;

2!2 = & k*.

1 4

Hence the required area is 7S k

s It is to be noted that the area DBCsince it is that of a right triangle,

could have been found by the formulas ^of plane geometry ;

for the altitude

DC = | and the base DB = l~^-~,2 z 2 4 4

and hence the area =16

Or we may construct the element

of area as shown in Fig 88

Then, if x1 and z2 are the abscissas

respectively of Paand P

t ,

.

2 4 3k 48

Ex. 3, Let the ellipse of Ex 1 be represented by the equations

x = a cos<f>, y = l sin <.

Using the same element of area, and expressing y and dx in terms of

we have , , ,, . ....dA = (b sin

</>) ( a sin

As x vanes from to a, </varies from to

;

A

hence =4 ft/O

=- 4

EEVIEW PROBLEMS 227

It is evident from foimula (1), 23, that the sign of a definite integralis changed by interchanging the limits. Hence

A =

irab

Ex. 4. Find the volume of the ring solid generated by revolving a

ciicle of radius a about an axis in its plane b units fiorn its center (b > a)

Take the axis of i evolution Yas OY (Fig. 89) and the line

through the center as OX. Thenthe equation of the ciicle is K

(r-&)2 + y2 =a2

.' M"--A straight line parallel to OX

meets the circle in two points .

Pv where a,=xi

b Va2y*, and

P2 , where x = xz = b + Va2ij\

A section of the requned solid

made by a plane through P1P2

perpendicular to OY is bounded

by two concentric circles with

radii MPl= x

vand MPz

= xz respectively. Hence, if dV denotes the

PIG. 89

element of volume, dy=

The summation extends from the point L, wheie y = a, to the point If,

where y = a. On account of symmetry, however, we may take twice the

integral from y = to y = a. Hence

V- 2 CJo

PIG. 90

Ex. 5. Find the pressure on a

parabolic segment, with base 2 b and

altitude a, submerged so that its

base is in the surface of the liquid

and its axis is vertical

Let RQ.C (Fig. 90) be the parabolic segment, and let CB be drawn

through the vertex C of the segment perpendicular to RQ, an the surface of

the liquid. According to the data, RQ = 2 ft, CB-a. Draw LN parallel

to Tti, and on LN as a base construct an element of area, dA. Let

CM=x.

228 APPLICATIONS

Then dA *=(LN)ilje.

r> , *But, from 30,

whencea

2/; Iand therefore dA = - <to.

*

The depth of Zi^V below the surface of the liquid is CH (W -,r ;

hence, if 10 is the weight of a unit volume of tho liquid,

tip sz 1 $* (a x) wdxt

a*

, .. ra 2biii i, x ,

and P -{ r j?

1

(-

.r)of.c

Jo

EXERCISES

1. Find the area of an arch of tho curve y = sin *.

w / * _ > \

2. Find the area bounded by tho catenary y = A v'"Hh " % fcliw

axis ofa?, and the lines r = /*.

3. Find the area included between tho curve ?/= -

..

*

. ., and.... *7 r + -1 rt^its asymptote.

4. Find the area of one of the closed figurus boundtul by Uit*

curves if = 16 a? and ?/2 = sc

8.

6. Find the area bounded by the curve 2/asss2(ce 1) and tilio

Iine2a; 3?/ =6. Find the area between the axis of SB and ono arch of tht

cycloid a; = a(j!> sin ^), y = a(l COH 0).

7. Find the volume of tho solid generated by revolving about OYthe plane surface bounded by OY and the curve* a$ -j- ^ s 3,

8. Any section of a certain solid made by a piano pM'pwulinulnr to

OX is an isosceles triangle with tho ends of its busa r8tint? on Ihw

ellipse 1 4- '4= 1 and its altitude equal to tho distance of tho plane

from the center of the ellipse. Find the total volume o tho Bulid,

9, Find the volume of the solid formed by revolving about tho

line 2 y + a the area bounded by 0110 arch of tho curve y m sin as

and the axis of x.

REVIEW PROBLEMS 229

10. Find the volume of the solid formed by revolving about the

line y + a = the area bounded by the circle a8 + t/2 = a2

.

11. Find the volume of the solid formed by revolving about the

line x a the area bounded by that line and the curve ay2 = x3

.

12. A right circular cone with vertical angle 60 has its vertex at

the center of a sphere of radius a Find the volume of the portionof the sphere included in the cone.

13. A trough 2 ft. deep and 2 ft. broad at the top has semiellipticalends If it is full of water, find the pressure on one end.

14. A parabolic segment with base 18 and altitude 6 is submergedso that its base is horizontal, its axis vertical, and its vertex m the

surface of the liquid Find the total pressure.

15. A pond of 15 ft. depth is crossed by a roadway with vertical

sides. A culvert, whose cross section is in the form of a parabolic

segment with horizontal base on a level with the bottom of the

pond, runs under the road. Assuming that the base of the parabolic

segment is 4 ft. and its altitude is 3 ft., find the total pressure on

the bulkhead which temporarily closes the culvert.

1. Find the pressure on a board whose boundary consists of a

straight line and one arch of a sine curve, submerged so that the

board is vertical and the straight line is in the surface of the water.

78. Infinite limits or integrand. There are cases m which it

may seem to be necessary to use infinity for one or both, of the

limits of a definite integral, or in which the integrand becomes

infinite. We shall restrict the discussion of these cases to the

solution of the following illustrative examples :

Ex. 1. Find the area bounded by the curve y = -^ (Fig. 91), the axis of

x, and the ordinate x = 1.x

It is seen that the curve has the axis of x as an asymptote ; and hence,

strictly speaking, the required area is not completely ^bounded, since the curve and its asymptote do not

intersect. Accordingly, in Fig. 91, let OM=l and

0JV = ft (& > 1) and draw the ordinates MP and

'

If the value of b is increased, the boundary line NQ, moves to the right;

and the greater b becomes, the nearer the area approaches unity.

230 APPLICATIONS

We may, accordingly, define, the area boundt'd by the eune, Ihe u\i* of r,

and the oidinate .r = 1 as tlw hunt of the area Mj\Ql*\w It tm'tcnst's indeti

nitely, and denoto it by the symbol

-'f = Lim fft

'('' .- 1.

Ex. & Find the, area bounded by the eurvt'//

-

^(KU?. tiiJ), tlw

axis of X) and the oidinatoH JK and u' -- .v "" " '"

Sinco the line u = <t is an jwyinptntc' of Ihc eum*, //~~- * when i

fmthtiimoH1

,ilic area in not, ntneUy speukiuj-, hounded. We IIH

(\, lun

find the area bounded on thci ri^hl by the tirdinuto

a; = a Ji, where ft is a Htuall quantity, with the lesult

If A M), Hin~ l

a -A'

(I

7T

Hence wo may regard J as tlui vahu of th> area

required, and express it by tlw

if **(/"

"*~~* M""jj"

1

ii

*

Ftu. UiJ

Ex. 3. Find the valuo ofJ

'

"f,

Proceeding as in Ex, 1, w( plaoo

/ f(JG * . /** wJ*

/-

-7--a Lim

/ ,.

^i V/ h **<' i Vj

But

an expression which increases Indeflnifcely m * Voo } hrn^o th* givt*nintegral has no finite value,

We accoidmgly conclude that in ranli aawt w rntiut dftermiuo n limit,and that the problem has no solution if wt* cannot find a

79. Area in polar coordinates, Let (Fig, 98) he thn po! midOH the initial line of a system of polar noSniinatai (r, &}< nf>and 0% two fixed radius vectors for which 0*0 and Bmirespectively, and P& any curve for which the aquation Ift

Requirod the area JJOE

AREA IN POLAR COORDINATES 231

To construct the differential of area, dA, we divide the angleinto parts, dd. Let OP and OQ be any two consecutive

radius vectors ; then the angle POQ = d0. With as a centerand OP as a radius, we draw the arc of a circle, intersecting OQat R. The area of the sector

It is obvious that the re-

quired area is the limit of the

sum of the sectors as their

number is indefinitely in-

creased. Therefore we have

and

FIG. 93This result is unchanged

if Pl coincides with 0, but

in that case OPlmust be tangent to the curve. So also Pz may

coincide with 0.

Ex. 1. Find the area of one loop of the curve r = a sin 3 (Fig 65, 51)

As the loop is contained between the two tangents 0=0 and 6 = , the

required area is given by the equation8

= fJo

12

Ex. 2. Find the area bounded by the lines 0= ~ and 0=~t the curve~44r = 2 a cos 0, and the loop of the curve r = a cos 2 6 which is bisected bythe initial line.

Since the loop of the curve r = a cos 2 Q is tangent to the line OL

(Fig. 94), for which = - ~, and the line ON, for which 6 f it is evi-

dent that the required area can be found by obtaining the area OLMNO,bounded by the lines OL and ON and the curve r 2 a cos 6, and subtract-

ing from it the area of the loop. The area may also be found as follows :

Let OPjPg be any radius vector cutting the loop r = a cos 2 Q at Pt and

the curve r = 2 a cos 6 at Pa. Let OPl

= rtand 0P2 rv Draw the radius

232 APPLICATIONS

vector OQ^jj, making an angle dd with OP^P^ With OPland OPS

as-

radii and as a center, constiuct arcs of circles niti'iaoetouft O(^l^ at A\

and J22 respectively. Then the area ot the sector J\01fi IH ^ rfilQ and tho

area of the sector PZORS is ^ r| dd. We^y

may now take the area P1/>

3JBa/'

and have

Then ^ =/ __ 7T ,_

4

01, since the required area is symmetricalwith respect to the line OM, wo may place

(>2-*i)0. FIG, 04

From the curve r = 2 a cos0, we have rf = 4 a8 coss 0, and from the curve

r = a cos 2 0, we have ra = a2 cos2 2 0; so that finally

- aa cos2 2 ^) <W

e-?-'222 8

EXERCISES

1. Find the total area of the lemniscato ?'a= 2 2 cos 2 0,

2. Find the area of one loop of the curve r =s a sin n 0,

3. Find the total area of the cardioid r = a(l-f- COB 0).

4. Find the total area bounded by the curve ? a 6 + 3 <uw 0.

5. Find the area of the loop of the curve r* t a" cos 2 non 3which is bisected by the initial line.

6. Find the area bounded by the curves r = a cos 8 and r == .,

7. Find the total area bounded by the curve r = 3 + 2 COB 4 0.

8. Find the area bounded by the curve ?cosa^=sl and the

7T *lines = and =

75.

9. Find the area bounded by the curves r ^ 6 + 4 cos andr s= 4 cos

10. Findtheareaboundedbythecurvesr=acos0andra=:a8 cos2^

MEAN VALUE 233

B

80. Mean value of a function. Let f(x) be any function of xand let y =/() be represented by the curve AB (Fig. 95), where

OM=a and ON=b. Take the points M^ Mz, , Mn_ r so as

to divide distance MN into

n equal parts, each equal to

dx, and at the points M, M^ Mz ,

'-, Mn _ l

erect the ordinates

2/0' yt y> ' &.-! Tlien the

average, or mean, value of these

n ordinates is

M

This fraction is equal to

ndx, dx-\

b a

If n is indefinitely increased, this expression approaches as a

limit the value

This is evidently the mean value of an"infinite number "

of

values of the function /(x) taken at equal distances between

the values x = a and x = 6. It is called the mean vafote of the

function for that interval.

Graphically this value is the altitude of a rectangle with the

base MN which has the same area as MNBA which equals

//aWe see from the above discussion that the average of the

function y depends upon the variable x of which the equal

intervals dx were taken, and we say that the function was

averaged with respect to x. If the function can also be averaged

with respect to some other variable which is divided into equal

parts the result may be different. This is illustrated in the

examples which follow.

234 APPLICATIONS

Ex. 1. Find the mean velocity of a body falling from rest duimg the

tune ti if the velocity is averaged with respect to the tune.

Here we imagine the time from to ^ divided into equal intervals dt

and the velocities at the beginning of each interval averaged. Proceeding

as in the text, we find, since v gt, that the mean velocity equals

1

Since the velocity is gt^ when t = tv it appears that in this case the

mean velocity is half the final velocity.

Ex. 2 Find the mean velocity of a body falling from rest through a

distance s1 if the velocity is averaged with respect to the distance.

Here we imagine the distance from to s1divided into equal intervals

ds and the velocities at the beginning of each interval aveiaged Pro-

ceeding as in the text, we find, since v = V2 y$, that the mean velocity is

Since the velocity is v^2 gsv when s = sv we see that in this case the

mean velocity is two thirds the final velocity.

EXERCISES

1. Find the mean value of the lengths of the perpendicularsfrom a diameter of a semicircle to the circumference, assuming the

perpendiculars to be drawn at equal distances on the diameter.

2. Find the mean length of the perpendiculars drawn from the

circumference of a semicircle to its diameter, assuming the perpen-diculars to be drawn at equal distances on the circumference

3. Find the mean value of the ordmates of the curve y = sm x7T

between x = and x =-5- , assuming that the points at which the

ordmates are drawn are at equal distances on the axis of x.

4. The range of a projectile fired with an initial velocity VQand

v^an elevation a is sin 2 a. Find the mean range as a varies from

TT ff

to 7p averaging with respect to a.

6. Find the mean area of the plane sections of a right circular' cone of altitude h and radius a made by planes perpendicular to theaxis at equal distances apart.

LENGTH OF PLANE CURVE 235

6. In a sphere of radius a a series of right circular cones is

inscribed, the bases of which are perpendicular to a given diameter

at equidistant points. Find the mean volume of the cones

7. The angular velocity of a certain revolving wheel varies with

the time until at the end of 5 mm. it becomes constant and equal to

200 revolutions per minute If the wheel starts from rest, what is its

mean angular velocity with respect to the time during the interval

in which the angular velocity is variable ?

8. The formula connecting the pressure p in pounds per squareinch and the volume v in cubic inches of a certain gas is pv = 20.

Find the average pressure as the gas expands from 2^- cu in. to 5 cu. in.

9. Show that if y is a linear function ofaj,

the mean value of ywith respect to x is equal to one half the sum of the first and the

last value of y in the interval over which the average is taken.

81. Length, of a plane curve. To find the length of anycurve AB (Fig. 96), assume n 1 points, J?, J?, , -_ r be-

tween A and B and connect each pair of consecutive points bya straight line. The length of AB is

then defined as the limit of the sumof the lengths of the n chords AQ,

%%, ]%PV -, X^.iB as n is increased

without limit and the length of each

chord approaches zero as a limit. Bymeans of this definition we have already

shown ( 39 and 52) that"

FIG. 96

d8**^dy?+df (1)

in Cartesian coordinates, and

<2 = V,fra4-rW (2)in polar coordinates.

Hence we have s = I ^dx*+ dy* (3)

and s => C-Vdr*+ r*d6\ (4)

To evaluate either (3) or, (4) we must express one of the

variables involved in terms of. the cither, or both in terms of a

third* The limits of integration may then be determined.

236 APPLICATIONS

Ex. I. Find the length of the parabola y2 = kx from the vertex to the

point (a, 6).

From the equation of the parabola we find 2 ydy = kdx Hence formula

(3) becomes either

Either integral leads to the result

Ex. 2. Find the length of one arch of the cycloid

x = a(<jf> sm<), y = a (1 cos <).

We have dx a (1 cos <) dtp, dy a sin<j> d$ ;

whence, from (1), ds = a V2 2 cos <j>d<j>= 2a sin 2

rf<ji>.

/* 27r <iTherefore s 2 a |

sin d<& = 8 a./o 2

EXERCISES

v 1. Find the length of the curve 3y*(x I)8 from its point of

intersection with OX to the point (4, 3").

af - --^w 2. Find the length of the catenary y -~\e a + e ") from x =to ai = h.

v 3. Find the total length of the curve

4. Find the total length of the curve x = a cos8 ^>, y = a sin8<.

5. Find the length of the curve

x = a cos<f> -}- a<l> sin <, y = a sin < a< cos <,

from < = toj!

= 4 TT.

6. Find the length of the curve x e~*cos t, y= e-*sin#, between7T

the points for which t = and t = -^.

7. Find the length of the curve r = a cos4

j from the point onthe curve for which 6 ~ to the pole

8. Find the total length of the curve r = a (1-|- cos 5).

9. Show that the length of the logarithmic spiralrea between

any two points is proportional to the difference of the radius vectors.

of the- points,

WORK 237

82. Work. By definition the work done in moving a body

against a constant force is equal to the force multiplied by the

distance through which the body is moved. If the foot is taken

as the unit of distance and the pound is taken as the unit of

force, the unit of measure of work is called & foot-pound. Thus

the work done in lifting a weight of 25 Ib. through a distance

of 50 ft. is 1250 ft-lb.

Suppose now that a body is moved along 0JT(Fig. 97) from

A (x = a) to B (x = b") against a force which is not constant but

is a function of a;, expressed by /(). Let the line AB be

divided into intervals each equal A i j/W J> xto dx, and let one of these inter-

vals be MN, where OM x. Then

the force at the point M is /(a?), and if the force were con-

stantly equal to f(x) throughout the interval MN, the work

done in moving the body through MN would be/(x) dx. This

expression therefore represents approximately the work actually

done, and the approximation becomes more and more nearly

exact as MN is taken smaller and smaller. The work done in

moving from A to B is the limit of the sum of the terms f(x) dx

computed for all the intervals between A and B. Hence we have

and/>

= /

Jit

Ex. The force which resists the stretching of a spring is propor-

tional to the amount the spring has been already stietched. Foi a cer-

tain spring this force is known to be 10 Ib. when the spring has been

stretched & in. Find the work done in stretching the spring 1 in. from

its natural (unstretched) length.

If F is the force reqmied to stretch the spring through a distance x,

we have, from the statement of the problem,

and since F ~ 10 when x = %, we have k = 20. Therefore F= 20 x.

Reasoning as in the text, we have

w~C\Jo

238 APPLICATIONS

EXERCISES

1. A positive charge m of electricity is fixed at 0, The repulsion

on a unit charge at a distance x from is -3 Find tho work done33

m bringing a unit charge from infinity to a distance a from 0.

2. Assuming that the force required to strotdi a wiro from the*t*

length a to the length a + x is proportional to -'j and that a foixioif*

of 1 Ib stretches a certain wire 36 in in lungth to a length .0,'i in.

greater, find the work done in stretching that wiro fromM in, to 40 in.

3. A block slides along a straight line from O against a resistanceka?

equal to .a

? where 7c and a are constants and or, is the distance(K j" Cb

of the block from at any time. Find the work done in moving theblock from a distance a to a distance aV from 0.

4. Find the foot-pounds of work done in lifting to a height of20 ft above the top of a tank all the water contained in a full cylin-drical tank of radius 2 ft. and altitude 10 ft.

5. A bag containing originally 80 Ib. of sand is liftod through avertical distance of 8 ft. If the sand leaks out, at such a rato that whilethe bag is being lifted, the number of pounds) of sand lost is wjual to aconstant times the square of the number of foot through which the baghas been lifted, and a total of 20 Ib. of sand i& lost during the lifting,find the number of foot-pounds of work done in lifting tho bag,

6. A body moves in a straight line according to the formula c ef,where x is the distance traversed in a time t. If tho resistance of theair is proportional to the square of tho velocity, find tho work doneagainst the resistance of the air as the body moves from != to ft* a.

7. Assuming that above the surface of the earth the form* of theearth's attraction varies inversely as tho square of the diBtanca fromthe earth's center, find the work done in moving a weight of w poundsfrom the surface of the earth to a distance a miles above the surfaca.

8. A wire carrying an electric current of magnitude C is bentinto a circle of radius a. The force exerted by the current upon aunit magnetic pole at a distance a from the center of the circle in astraight line perpendicular to the plane of the circle is known to be27rCas

(a?+^' done in bringin8 a ttnit magnetic pole from

infinity to the center of the circle along the line just mentioned,

GENERAL EXEKClSES 239

9. A piston is free to slide in a cylinder of cross section S Theforce acting on the piston is pS, where p is the pressure of the gasin the cylinder, and is 7.7 Ib. per square inch when the volume v is

2 5 cu in. Find the work done as the volume changes from 2 cu. in. to

6 cu. in., according as the law connecting p and v is (1) pv k or

(2)X* = *GENERAL EXERCISES

1. Find the area of the sector of the ellipse 4oj2+9y2 =36cut out of the first quadrant by the axis of x and the line 2 y = x.

2. Find the area of each of the two parts into which the area of

the circle a;3 + if= 36 is divided by the curve y

2 = a;8.

3. Find the area bounded by the hyperbola xy = 12 and the

straight line x + y 8 = 0.

4. Find the area bounded by the parabola a;2=4 ay and the

8 a8

5. Find the area of the loop of the curve ay2 = (x a) (x 2 a)

z.

6. Find the area of the two parts into which the loop of the

curve y* = cc2

(4 a;)is divided by the line x y = 0.

7. Find the area bounded by the curve ary* + a262 = a22/a and its

asymptotes.

8. Find the area bounded by the curve 2/2

(aj

a + &2

)= <&& and its

asymptotes.

9. Find the area bounded by the curve a? = a cos 6, y = b sin8

10. Find the area inclosed by the curve x = a cos80, y = a sin8

0.

11. Two parabolas have a common vertex and a common axis, but

lie in perpendicular planes. An ellipse moves with its plane perpen-dicular to the axis and with the ends of its axes on the parabolas.

Find the volume generated when the ellipse has moved a distance

h from the common vertex of the parabolas.

12. Find the volume of the solid formed by revolving about the line

a;= 4 the figure bounded by the parabola y*= 4x and the line x = 1.

13. A right circular cylinder of radius a, is intersected by two

planes, the first of which is perpendicular to the axis of the cylinder

and the second of which makes an angle with the first. Find the

volume of the portion of the cylinder included between these two

planes if their line of intersection is tangent to the circle cut from

the cylinder by the first plane.

240 APPLICATIONS

fn O

., w. u^ U.UU.UJLU w*.^c*Uw ~~ ~ + y* a? as a base,

an isosceles triangle is constructed with its altitude equal to the

ordinate and its plane perpendicular to the plane of the curve. Find

the volume generated as the triangle moves from x = a to x = a.

15. Find the volume of the solid generated by revolving about

8 a8

the line OF the figure bounded by the curve y = 2 . and the

line y o,.

16. Find the volume of the solid formed by revolving about the

line x = 2 the plane area bounded by that line, the parabola yz = 3x,

and the lines y = 3.

17. Find the volume formed by revolving about the line x = 2

the plane figure bounded by the curve y3 = 4 (2 a;)

and the axis of y.

18. The sections of a solid made by planes perpendicular to OFaie circles with one diameter extending from the curve ^ 4 a; to

the curve y2 = 4 4 x. Find the volume of the solid between the

points of intersection of the curves.

19. The area bounded by the circle x* + $ 2 ax = is revolved

about OX, forming a solid sphere Find the volume of the two parts

into which the sphere is divided by the surface formed by revolving

3.8

the curve y2 = about OA'y 2a x

20. Find the volume of the two solids formed by revolving about

Y the areas bounded by the curves ar8

-f #2 = 5 and

g/

2 = 4 ai.

21. Find the volume of the solid formed by revolving about OXthe area bounded by OX, the lines x = and x == a, and the curve

z

y = x + aea.

22. The three straight lines OA, OB, and OC determine two planes

which intersect at right angles in OA. The angle A OB is 45 and the

angle AOC is 60. The section of a certain solid made by any plane

perpendicular to OA is a quadrant of an ellipse, the center of the

ellipse being in OA, an end of an axis of the ellipse being in OB, and

an end of the other axis of the ellipse being m OC. Find the volume

of this solid between the point and a plane perpendicular to OA at

a distance of two units from 0.

23. The section of a solid made by any plane perpendicular to OXis a rectangle of dimensions a? and sm x, x being the distance of the

plane from O. Find the volume of this solid included between the

planes for which x = and x = IT.

GENERAL EXEECISES 241

24. An oil tank is in the form of a horizontal cylinder the endsof which are circles 4 ft. in diameter. The tank is full of oil, which

weighs 50 Ib. per cubic foot Calculate the pressure on one end of

the tank.

25. The gasoline tank of an automobile is in the form of a hon-zontal cylinder the ends of which are plane ellipses 20 in high and10 in broad. Assuming w as the weight of a cubic inch of gasoline,find the pressure on one end of the tank when the gasoline is 15 in

deep.

26. A horizontal gutter is U-shaped, a semicircle of radius 3 in,

surmounted by a rectangle 6 in wide by 4 in. deep If the gutter is

full of water and a board is placed across the end, how much pressureis exerted on the board ?

27. The end of a horizontal gutter is in the form of a semicircle

of 3 in. radius, the diameter of the semicircle being at the top andhorizontal. The gutter receives water from a roof 50 ft above the

top of the gutter. If the pipe leading from the roof to the gutter is

full, what is the pressure on a board closing the end of the gutter ?

28. A circular water main has a diameter of 5 ft. One end is closed

by a bulkhead, and the other is connected with a reservoir in whichthe surface of the water is 20 ft above the center of the bulkhead

Find the total pressure on the bulkhead.

29. Find the area of a loop of the curve ?*a= a? sin n6.

30. Find the area swept over by a radius vector of the curveIT

r a tan as $ changes from to4

431. Find the area inclosed by the curve r = H .. and the

.

J 1 cos $

curve r = H ,

1 + cos 6

32. Find the area bounded by the circles r=a cos 6 and r= a, sin 6.

33. Find the area cut off from one loop of the curve r2= 2 a2 sin 2 6

by the circle r a

34. Find the area of the segment of the cardioid r = a (1+ cos 0)

cut off by a straight line perpendicular to the initial line at a dis-

tance | a from the origin 0.

35. Find the area cut off from a loop of the curve r = a sm 3 6 by

. , . ,

the circle r s=

242 APPLICATIONS

36. Find the area cut off from the lemniscate r2 = 2 a2cos 2 by

_/*>

the straight line r cos =75

'

37. Find each of the three areas bounded by the curves r = a

and r = a (1+ sin 0). 8

38. Find the mean height of the curve y 2. 2 between the

lines x = 2a and aj = 2 a.

39. A particle describes a simple harmonic motion defined by the

(792i?/

\

~9~)

during a complete vibration is half the maximum kinetic energy iL

the average is taken with respect to the time.

40. In the motion defined in Ex 39 what will be the ratio of

the mean kinetic energy during a complete vibration to the maxi-

mum kinetic energy, if the average is taken with respect to the

space traversed?

41. A quantity of steam expands according to the Iawj0v08 =; 2000,

p being the pressure in pounds absolute per square foot Find the

average pressure as the volume v increases from 1 cu. ft. to 5 cu. ft.

a242. Find the length of the curve y = a In

2. from the origin

a ^to the point for which x = ~

43. Find the length of the curve y = Li'x

- between the pointsfor which x =sl and x = 2 respectively.

~~

44. Find the total length of the curve x = a cos8<, y = b sin*<.a

45. Find the total length of the curve r = a sin8

^

46. Find the length of the spiral r = aO from the pole to the end

of the first revolution.JA

47. If a center of force attracts with a magnitude equal to r>

SC#

where x is the distance of the body from the center, how much workwill be done in moving the body in a straight line away from the

center, from a distance a to a distance 8 a from the center ?

48. A body is moved along a straight line toward a center of

force which repels with a magnitude equal to 7ccc when the bodyis at a distance x from the center. How much work will be donein moving the body from a distance 2 a to a distance a from the

ter?

GENERAL EXERCISES 243

49. A central force attracts a body at a distance x from the center7c

by an amount -j Find the work done in moving the body directly

away from the center from a distance a to the distance 2 a.

50. How much work is done against hydrostatic pressure in rais-

ing a plate 2 ft. square from a depth of 20 ft. to the surface of the

water, if it is kept at all times parallel to the surface of the water ?

51. A spherical bag of radius 5 in. contains gas at a pressure

equal to 15 Ib. per square inch. Assuming that the pressure is in-

versely proportional to the volume occupied by the gas, find the

work required to compress the bag into a sphere of radius 4 in.

CHAPTER XI

REPEATED INTEGRATION

83. Double integrals. The symbol

nvt

f(x, y) dxdy, (1)_ 4

in which a and b are constants and y^ and y^ are either con-

stants or functions of x, indicates that two integrations are to

be carried out in succession. The first integral to be evaluated is

where x and dx are to be held constant. The result is a func-

tion of x only, multiplied by dx; let us say, for convenience,

;)dx.

The second integral to be evaluated is, then,

F(x) dx,

which is of the familiar type.

Similarly, the symbol6 /*xs

f(x, y) dydx, (2)

where a and b are constants and ^ and #gare either constants

or functions of #, indicates first the integration

in which y and dy are handled as constants, and afterwards

integration with respect to y between the limits a and b.

244

ftn 2

Ex. 1. Evaluate / I xythdy.

The first integral is

j scydxily

The second integration is

DOUBLE INTEGRALS 245

s /.a

Ex. 2. Evaluate f f"'

O 2 + y*)<lr.diiJo Ji-x ^ '

The ni*st integration is

The second integration is

Ex.3. Evaluate f^" f**jpdyt?jc.Jo i/o

The first integration is

v3 r n j/a

(u

y*<lytlx=

?/

2^/y* = /-.//.

1/0 L Jo <l'

The second integration is

r2a^y= rir

a= ! a4

Jo 4a ''

LaOoJo 6

A definite integral in one variable has been shown lo be the

limit of a sum, from which we infer that formula (1) involves

first the determination of the limit of a sum with respect to;/,

followed by the determination of the limit of a sum with respectto x. The application of the double integral comes from it

interpretation as the limit of a double summation,How, such forms arise in practice will be illustrated in the

following sections,

EXERCISES

Find the values of the following integrals ;

a"%// 8 r*

^dydx, 3./ /

Ji J,,

n\xydoady. 4.

'

246

5.

REPEATED INTEGRATION

dydx

8.

/ /-as

|I

c/O ^/o

rf rJo /o r8

a(l + coi S)

? sin Odddr.

r*d$dr.

r cos 6dOdr

84. Area as a double integral. Let it be required to find an

area (such as is shown in Fig. 98) hounded by two curves, with

the equations y^~f^(x) and y2=/2 () intersecting in points for

which x a and z=b respectively. Let the plane be divided into

rectangles by straight lines parallel to OX and OT respectively.Then the area of one such rectangle is

where dx is the distance between two consecutive lines parallelto OY3 and. where dy is the distance between, tw,o consecutive

lines parallel to OX. The sum of the rectangles which are either

AKEA AS DOUBLE INTEGRAL 247

wholly or partially within the required area will be an approx-imation to the required area, but only an approximation, because

the rectangles will extend partially outside the area. We assume

as evident, however, that the sum thus found becomes more

nearly equal to the required area as the number of rectanglesbecomes larger and dx and dy smaller. Hence we say that the

required area is the limit of the sum of the terms dxdy.The summation must be so carried out as to include every

rectangle once and only once. To do this systematically we

begin with any rectangle in the interior, such as PQJRS, and addfirst those rectangles which lie in the vertical column with it.

That is, we take the limit of the sum of dxdy, with x and dx

constant and y varying from y^f^(x) to 2/2=/fl (X). This is

indicated by the symbol

/"vt/.

x = [/2 (x) -/,()] dx. (2)

This is the area of the strip TUVW. We are now to take

the limit of the sum of all such strips as dx approaches zero

and x varies from a to 5.

We have then

^ <=/"<>,-^^ = fV, (x) -/x <V>] dx. (3)Jo. Ja

If we put together what we have done, we see that we have

n s

dxdy. (4)_.

This discussion enables us to express the area as a double

integral. It does not, however, give us any more convenient wayto compute the area than that found in Chapter III, for the result

(2) is simply what we may write down at once for the area of a

vertical strip (see Ex. 3, 28).If it should be more convenient first to find the area of a

horizontal strip, we may write

n* dy dx. (5)

248 BEPEATED INTEGRATION

Consider a similar problem in polar coordinates. Let an

area, as in Fig. 99, be bounded by two curves r^/X^Oand r

2 =/s (0), and let the values of corresponding to the

points B and beland

a respectively. The plane may be

divided into four-sided figures by circles with centers at and

straight lines radiating

from 0. Let the angle

between two consecu-

tive radii be d6 and

the distance between

two consecutive circles

be dr. We want first

the area of one of the

quadrilaterals such as

PQRS. Here OP = r,

= dr, and the anglePOS=d0. By geom-

etry the area of the

sectorPOS=^r2d0 andA

the area of the sector FIG. 99

therefore PQRS= J (r + dr)*d6-

J r*d0 = rdrd0 + % (dr)*d0.Now as dr and d0 approach zero as a limit the ratio of the second

term in this expression to the first term also approaches zero,

since this ratio involves the factor dr. It may be shown that the

second term does not affect the limit of the sum of the expression,and we are therefore justified in writing as the differential of area

dA = rd0dr. (6)

The required area is the limit of the sum of these differ-

entials. To find it we first take the limit of the sura of the

quadrilaterals, such as PQRS, which lie in the same sector UO V.

That is, we integrate rd0dr, holding and dd constant and al-

lowing r to vary fromr^ to r. We have

/r. (7)

which is the area of the strip TUVW.

CENTER OF GRAVITY 249

Finally we take the limit of the sum of the areas of all such

strips in the required area and have

(8)

If we put together what we have done, we may write

dOdr. 9/*s /"a= I I r

t/0t J>i

It is clear that this formula leads to nothing which has not

been obtained in 79, but it is convenient sometimes to have

the expression (9).

85. Center of gravity. It is shown in mechanics that the cen-

ter of gravity of n particles of masses m^ w2 , -, mn lying m a

plane at points whose coordinates are^, yx), (xz , /2), , (#, y^)

respectively is given by the formulas

=

ss

\-mn

This is the point through which the resultant of the weightsof the particles always passes, no matter how the particles are

placed with respect to the direction of the earth's attraction,

We now wish to extend formulas (1) so that they may be

applied to physical bodies in which the number of particles maybe said to be infinite. For that purpose we divide the bodyinto n elementary portions such that the mass of each may be

considered as concentrated at a point (or, y). Then, if m is the

total mass of the body, the mass of each element is dm. We have

then to replace the m's of formula (1) by dm and to take the

limit of the sums involved in (1) as the number n is indefinitely

increased and the elements of mass become indefinitely small.

There result the general formulas

/ xdm I ydm-X= J , 9,J . (2)

I dm I dm

250 REPEATED INTEGRATION

To apply these formulas we consider first a slender wire so

fine and so placed that it may be represented by a plane curve.

More strictly speaking, the curve may be taken as the mathe-

matical line which runs through the center of the physical wire.

Let the curve be divided into elements of length ds. Then,if c is the area of the cross section of the wire and D is its

density, the mass of an element of the wire is Dads. For con-

venience we place DC = p and write

where p is a constant. If this is substituted in (2), the constant

p may be taken out of the integrals and canceled, and the

result may be written in the form

sx = ixds, sy =I

yds, (3)

where a on the left of the equations is the total length of the

curve. These formulas give the center of gravity of a plane curve.

Ex. 1. Find the center of gravity of one fourth of the circumference

of a circle of radius a.

Here we know that the total length is

\ rrax = Cxds, vay = Cyds.

To integrate, it is convenient to in-

troduce the central angle <j> (Fig 100),whence x = a cos <j>,y

= a sin<f>,

ds = ad<j>.

7T

Then |irox= l*a*cos<f>dd>,Jo

,

whence

we,so that, from (3), we have

2aFIG. 100

Consider next a thin plate, which may be represented by a

plane area. Strictly speaking, the area is that of a section

through the middle of the plate. If t is the thickness of the

plate and _D its density, the mass of an element of the platewith the area AA is DtdA. For convenience we place Dt = pand write

dm pdA,

CENTER OF GEAVITY 251

where p is a constant. If this is substituted in (2) and the p's

are canceled, we have

i = I xdA, AyI ydA, (4)

where A is the total area. These formulas give the center of

gravity of a plane area,

Ex. 2. Find the center of gravity of the area bounded by the parabola

y* = kx, the axis of x, and the ordmate through the point (a, b) of the

parabola (Fig 101)

We place dA= dxdy in (4) and have

A js = CCxdx dy, Ay =Cfy dx dy.

To evaluate, we choose the element dxdyinside the area in a general position, and first

sum with respect to y along a vertical strip.

We shall denote by ylthe value of y on the _ i .

parabola, to distinguish it from the general^

values of y inside the area The first integra-

tion gives us, therefore, respectively

CVi

xdxdy = xy1 dx and C l

ydxdy \y* dx,

so that we have Ax Cxy^dx, Ay = C^y* dx-J /

On examination of these results we see that each contains the factor

y^dx (which is the area ( 22) of an elementary vertical strip), multiplied

respectively by a? and ^ yv which are the coordinates of the middle point of

the ordmate yr These results are the same as if we had taken dA = ytdx in

the general foimula (4), and had taken the point (x, y) at which the mass

of dA is concentrated as (z, yx), which is in the limit the middle point of

dA. In fact this is often done in computing centers of gravity of plane

areas, and the first integration is thus avoided.

Kow, from the equation of the parabola y%= kx, and to complete the

integration, we have to substitute this value for yt and integrate with

respect to x from x = to x = a We have

AS =

* from the

/** yfco"

Ay i i kxdx <-

of the curve, k = and, by 23, A ~ -<

fl O

252 REPEATED INTEGRATION

Substituting these values and ieducing, we have finally

x = | a, y=&bIn solving this problem we have carried out the successive mtegratioii.s

separately, in oider to show clearly just what has been done. If now wecollect all this into a double integral, we have

Ax=fa

f ''xdxdy. /ly=f"f 'ytlxda.Jo Jo Jo Jo

Ex. 3. Find the center of gravity of a sextant of a circle of radum a

To solve this problem it is convenient to place the sextant so that theaxis of v bisects it (Fig. 102) and to use ._

polar coordinates.

From the symmetry of the figure the

centei of gravity lies on OX, so that we

may write at once y = 0. To find a take ran element of area ? dddr in polar cooi-

dinates and place x = r cos We have

then, from (4),a

/"r ^

/r2 cos 6dddr,

rr Jo~6

where A = lira?, one sixth the area of aciicle In the first integration 6 and dOare constant, and the summation takes

place along a line radiating from with

r varying from to a The angle then varies from - - to ~ , and thus theentire area is covered The solution is as follows .

6

liraPx =J8 a3 cos Odd

whence

6

_ 2a

Consider now a solid of revolution formed by revolving theplane area (Fig. 103) ABCD about Y as an axis. It is assumedthat the equation of the curve CD is given. It is evident fromsymmetry that the center of gravity of the solid lies on OY, sothat we have to find only y.

Let dFbe any element of volume. Then dm = pdV, where p isthe density and is assumed constant. Substituting in (2), we have

(5)

CENTER OF GRAVITY 253

Let the solid be divided into thin slices perpendicular to

as was done in 26, and let the summation first take place over

one of these slices. In this summa-tion y is constant, and the result

oL the summation is therefore ytimes the volume of the slice. It is

therefore u(^r'^^y)- We have nowto extend the summation over all

the slices. This gives the result

r b

Vy = / ir^ydy, (G)/

TIG. 103

where OA = a and 07? = &.

It is to be noticed that this result

is what we obtain if we interpret

dm in (2) as the mass of the slice and consider it concentrated

at the middle point of one base of the slice.

Ex. 4. Find the centei of gravity of a light cucular cone of altitude 6

and radius (Fig. 104)

This is a solid of revolution formed by revolv-

ing a light triangle about OF. However, the

equation of a straight line noed not be used, as

X CL

similar triangles are simpler. We have - = ->

y b

whence x -y. The volume V is known to beb

J 7r 86. Therefore, from (G), we have

f*mP aj=J

- fdy =

whence FIG. 104

EXERCISES

1. Show that the center of gravity of a semicircunifereuce ofo a

radius a lies at a distance of from the center of the circle on7T

the radius which bisects the semicircumference.

2, Show that the center of gravity of a circular arc which subtends

an. angle <x at the center of a circle of radius a lies at a distance sin ~

from the center of the circle on the radius which, bisects the arc,

a

254 EEPEATED INTEGRATION-

3. A wire hangs so as to form the catenary y~

(/* -1- ").

Find the center of gravity of the piece of the curve between Iho

points for which x = and x = a.

4. Find the center of gravity of the arc of the cycloid

x = a(<[> sm$), y = a(l coB<f>), between the first two sharp

points.

5. Find the center of gravity of a parabolic segment of base, 2 b

and altitude a.

6. Find the center of gravity of a quadrant of the area of a circle.

7. Find the center of gravity of a triangle.

8. Find the center of gravity of the area bounded by the ourvo

y = sin x and the axis of x between x = and x = IT.

9. Find the center of gravity of the plane area bounded by tho

two parabolas y*= 20 x and a?= 20 y.

10. Find the center of gravity of a figure which is composed of

a rectangle of base 2 a and altitude I surmounted by a semicircle

of radius a

11. Find the center of gravity of the area bounded by the Ihvst

arch of the cycloid (Ex. 4) and the axis of aj.

12. Show that the center of gravity of a sector of a circle lies at4 a a

a distance ~ sin- from the vertex of the sector on a line bisecting

the angle of the sector, where a is the angle and a the radius.

13. Find the center of gravity of the area bounded by the cardioid

14. Find the center of gravity of the area bounded by the curver = 2 cos &+ 3.

15. Find the center of gravity of a solid hemisphere.16. Find the center of gravity of a solid formed by revolving

about its altitude a parabolic segment of base 26 and altitude a.

17. Find the center of gravity of the solid formed by revolvingabout OF the plane figure bounded by the parabola y*** fee, the axisof y, and the line y = k.

18. Find the center of gravity of the solid bounded by the sur-faces of a right circular cone and a hemisphere of radius & with thebase of the cone coinciding with the base of the hemisphere and thevertex of the cone in the surface of the hemisphere

CENTER OF GRAVITY 255

86. Center of gravity of a composite area. In finding the cen-

ter of gravity of a body which is composed of several parts the

following theorem is useful:

If a body of mass M is composed of several parts of masses

M^ Mf -,Mn, and if the centers of gravity of these parts are

respectively (xf ^), (za , /

?), .,(#w, #), then the center of grav-

ity of the composite body is yiven by the formulas

+ J

C ;

We shall prove the theorem for the % coordinate. The prooffor y is the same.

By 85 we have, for the original body,

MX - fxdm,'

(2)

where the integration is to be taken over all the partial masses

Mv Jfa , ,

Mn into which the body is divided. But we have also

where the subscripts indicate that the integration in each case

is restricted to one of the several bodies. But formula (2) can

be written r C CMX = / x^m^ I x^dm^-}- + / ^dmn ;

and, by substitution from (3), the theorem is proved,

x. Find the center of gravity of an area bounded by two circles one

of which is completely inside the other.

Let the two circles be placed aa in Fig. 105, where the center of the

larger circle of radius a is at the origin, and the center of the smaller

circle of radius o is on the axis of $ at a distance o from the origin*

256 REPEATED 1NTE( 1 RATU )K

The area which can be considered as composed of two ]>jut.n i that of

the larger circle, the two paits being, fust, the smaller eirde and, wroucl, UN-

irregular ring whose centei of gravity

we wish to find Now the center of

gravity of a circle is known to be at

its center, Theiefore, in the formula

of the theoiem, we know (T, J/),which

is on the left of the equation, to bo

(0, 0), and (~cv yj to be (c, 0), and wish

to find (72 , T/z).

Since we are dealing with aieas,

we take the masses to be equal to the

ai eas, and have, accordingly, M = iraz

(the mass of the larger ciiclc), M = irbz

(the mass of tho smallei cncle), and

My,= if

(a

fta) (the mass of tho ring).

Substituting in the formula, we havo jor>

= IT/A- + TT (<r- A-

1

) ra ;

whence, by solving for ,ra

It is unnecessary to find yi} since, by Hyiniuctiy, Mi coutiM of

lies on OXEXERCISES

1. Show that if there are only two component massesfl/^

mul.JHf^

in formulas (1) of the theorem, the center of gravity of tho oomjxmtomass lies on the line connecting the centers of gravity of tho com-

ponent masses at such a point as to divide that hue into sogmtmlHinversely proportional to the masses.

2. Prove that if a mass M^ -with center of gravity (.r^ //j)haw <m(,

out of it a mass Mzwith center of gravity (i?9, ya),

tho c-oiiUu1 of gravityof the remaining mass is

Mi-Mt J/t

- Mt

and rsaro tangtnit oxtonuJly. Find3. Two circles of radii

their center of gravity

4. Find the center of gravity of a hemispherical hol3 boundodby two concentnc hemispheres of radii

r^and r^

5. Place r9= ^ + A^ m Ex. 4, let A?' approach xero, and thus iind

the center of gravity of a hemispherical surface.

CENTER OF GRAVITY 25Y

6. Find the center of gravity of a hollow right circular cone

bounded by two parallel conical surfaces of altitudes 7^ and A2

respectively and with their bases in the same plane.

7. Place \= /^-H A/i in Ex 6, let A/t approach zero, and thus

find the center of gravity of a conical surface

8. Find the center of gravity of a carpenter's square each aim

o which is 15 in on its outer edge and 2 in wide

9. From a square of edge 8 in a quadrant of a circle is cut out,

the center of the quadrant being at a corner of the squaie and the

radius of the quadrant being 4 in. Find the center of gravity of the

figure remaining

10. Two iron balls of radius 4m and 6 in respectively are con-

nouted by an iron rod of length 1 in. Assuming that the rod is a

cylinder o radius 1 in., find the center of gravity of the system.

11. A cubical pedestal of side 4 ft. is surmounted by a sphere of

radius 2 ft Find the center of giavity of the system, assuming

that the sphere rests on the middle point of the top of the pedestal.

87. Theorems. The foliowing theorems involving the center

oi' gravity may often ho used to ad-Q

vantage in finding pressures, volumes x

c>I solids of revolution, or areas of

surfaces of revolution.

I. The total pressure on a plane sur-

faae immersed in liquid in a vertical

position is equal to the area of the sur-

face multiplied ly tlw pressure at its

center of gravity.

Let the area he placed a in Fig 106, ETO< 106

where the axis of K is in the surface

of the liquid and where the axis of y is measured downward.

Then, by 25, -

/* I yOpa- fl!i)^> '*'

which may be written IIH a double integral in the form

p = CCwydydx => w nydxdy, (2)

258 BEPEATED INTEGRATION

In fact, this may be written down directly, since the pressure on

a small rectangle dxdy is its area, dxdy, times its depth, T/, times w.

Moreover, from 85, we have

Ay --CCydxdy. (3)

By comparison of (2) and (3) we have

P = wyA.

But wyis the pressure at the center of gravity, and the theorem

is proved for areas of the above general shape. If the area is

not of this shape, it may be divided into such areas, and the

theorem may be proved with the aid

of the theorem of 86.

Ex.1. A circular bulkhead which closes y="b

the outlet of a reservoir has a radius 8 ft ,

and its center is 12 ft below the surface of

the water. Find the total pressure on it

Here A = 9 it and the depth of the center

of gravity is 12. Therefore

P = 108 irw = -V-Trtons = 10.6 tons.

II. The volume generated by revolving

a plane area about an axis in its planenot intersecting the area is equal to the area of the figure multiplied

by the circumference of the circle described by its center of gravity.

To prove this take an area as in Fig. 107. Then, by 26, if

V is the volume generated by the revolution about OY,

(4)

which can be written as a double integral in the form

F=27rf C\dydx. (5)t/ i/,

By 85, J&= C C\dydx;Ja

u/jTj

and, by comparison of (4) and (5), we have

V~which was to be proved.

THEOREMS OF PAPPUS 259

Ex. 2. Find the volume of the nag surface formed by revolving about

an axis in its plane a circle of radius a whose center is at a distance c from

the axis, where c > a.

We know that A = -sra2 and that the center of gravity of the circle is

at the center of the circle and therefore describes a circumference of length

2 ire. Therefoie F = 2ir2ffl2c.

III. The area generated by revolving a plane curve about an

axis in its plane not intersecting the curve is equal to the length

of the curve multiplied by the circumference of the circle described

by its center of gravity.

To prove this we need a formula for the area of a surface of

revolution which has not been given. It may be shown that if

S is this area, then rS^Zirlxds. (6)

A rigorous proof of this will not be given here. However, the

- student may make the formula seem plausible by noticing that

an element ds of the curve will generate on the surface a belt

of width ds and length 2 irx. The product of length by breadth

may be taken as the area of the belt.

Moreover, by 85, we have

sx I xds; (7)

and comparing the two equations (6) and (7), we have

which was to be proved.

Ex. 3. Find the area of the ring surface described in Ex. 2.

We know that s - 2 TTO and that the center of gravity of a circumfer-

ence is at its center and therefore describes a circumference of length 2 ire

Therefore 5 = 4 ^ac.

Theorems II and III are known as the theorems of Pappus.

EXERCISES

1. Find by the theorems of Pappus the volume and the surface

of a sphere.

2, Find by the theorems of Pappus the volume and the laterax

surface of a right circular cone.

260 .REPEATED INTEGRATION

3. Find by the theorems of Pappus the volume generated by

revolving a parabolic segment about its altitude.

4. Find by the theorems of Pappus Iho volume generated

by revolving a parabolic segment about its base.

5. Find by the theorems of Pappus the volume geru>ratod by

revolving a parabolic segment about the tangent at its vortex

6. Find the volume and the surface generated by revolving a

square of side a about an axis in its piano perpendicular to one. of

its diagonals and at a distance b\b> ;= }from it,s center.

V V2/7. Find the volume and the area generated by revolving' a right

triangle with legs a and b about an axis in its plane ]>urullel to the

leg of length a on the opposite side from the hypotenuse and at adistance c from, the vertex of the right angle.

8. A circular water main has a diameter of 4 ft. One end IH

closed by a bulkhead, and the other is connected with a reservoirin which the surface of the water is 18 ft. above the center of thebulkhead Find the pressure on the bulkhead.

9. Find the pressure on an ellipse of semiaxos a. and I completelysubmerged, if the center of the ellipse is o units below the surface ofthe liquid.

10. Find the pressure on a semiellipse of semiaxes a and I (a> />)

submerged with the major axis in the surface of the liquid and theminor axis vertical.

11. Find the pressure on a parabolic segment submerged with thebase horizontal, the axis vertical, the vertex above the base, and thevertex c units below the surface of the liquid.

^

12. What is the effect on the pressure of a submerged vertical areain a reservoir if the level of the water in the reservoir is raised by c feet ?

88. Moment of inertia. The moment of inertia of a particleabout an axis is the product of its mass and the square of itsdistance from the axis. The moment of inertia of a numberof particles about the same axis is the sum of the moments ofinertia of the separate particles abont that axis. From thesedefinitions we may derive the moment of inertia of a thin plate.Let the surface of the plate be divided into elements oarea cU Then the mass of each element is pdA, where p is theproduct of the thickness of the plate and ita density. Lot M bo

MOMENT OF INEETIA 261

the distance of any point in the element from the axis about

which we wish the moment of inertia. Then the moment of

inertia of clement is approximately

I?p dA.

We say"approximately

"because not all points of the element

are exactly a distance R from the axis, as R is the distance ot

some one point in the element. However, the smaller the ele-

ment the more nearly can it be regarded as concentrated at one

point and the limit of the sum of all the elements, as their

size approaches zero and their number increases without limit,

is the moment of inertia of the plate. Hence, if I represents the

moment of inertia of the plate, we have

= f (1)

If in (1) we let/o= l, the resulting equation is

1= C&dA, (2)

where I is called the moment of inertia of the plane area. When

dA in (1) or (2) is replaced by dxdy or rdrdd, the double

sign of integration must be used.

Ex. 1. Find the moment of inertia of a rectangle of dimensions a and 6

about the side of length &

Let the rectangle be placed as in Fig. 108. Let it be divided up into

elements dA ~dfdg. Then a- is the distance of some point in an element

from OY. Hence, in (2), we have yJR SB x and dA = fatly. Therefore

y=b

We first sum the rectangles in

a vertical strip, w y ranges from

to &. We have

This is the moment of ineitia ofj.IG< IQ%

the strip MN, and might -have been

written down at once, since all points on the left-hand boundary of the

strip are at a distance x from OY and since the area of the strip is Idx

262 KEPEATED INTEGRATION

The second integration gives now

Jo

If, instead of asking for the moment of inertia of the area, we had asked

for that of a plate of metal of thickness t and density D, the above result

would be multiplied by p = Dt. But m that case the total mass M of the

plate is aab, so that we haveI \ Ma?

Ex. 2. Find the moment of inertia of the quadrant of an ellipse

r2 7/2

}.>L 1 (a > 6) about its major axis

If we take any element of area as dxtly, we find the distance of its

lower edge from the axis about which we wish the moment ot ineitia

to be y (Fig 109). Hence R = y and ^I

It will now be convenient to sum first

with respect to xt since each point of a

horizontal strip is at the same distance from

OX "We therefore write

J-ffffy**. FIG. 109

Now, indicating by a^ the abscissa of a point on the ellipse to distinguishit from the general x which is that of a point inside the ellipse, we have

for the first integration

For the second integration

/

To integrate, place y = I sin<j>

ThenTT

= afc8 f* s/o 16

If, instead of the area, we consider a thin plate of mass M, the aboveresult must be multiplied by p, where M=

irabp ,whence

The polar moment of inertia of a plane area is defined as the

moment of inertia of the area about an axis perpendicular to

its plane. This may also be called conveniently the moment

MOMENT OF INERTIA 263

of inertia with respect to the point in which the axis cuts the

plane of the area, for the distance of an element from the axisis simply its distance from that point. Thus we may speak,for example, of the polar moment of inertia with respect to

an axis through the origin perpendicular to the plane of au

area, or, more concisely, of the polar moment with respect to

the origin.

If the area is divided into elements dxdy, and one point in

the element has the coordinates (#, #), the distance of that

point from the origin is Vi?+]A That is, in (2), if we place(lA dxvly and J2

a=o;2+y2, we shall have the formula for the

polar moment of inertia with respect to the origin. Denotingthis by J

, we have

(3)

This integral may be split up into two integrals, giving

(4)

where the change in the order of the differentials in the two

integrals indicates the order in which the integration may be

most conveniently carried out.

The first integral in (4) is the moment of inertia about OYand may be denoted by /

; the second integral is the momentof inertia about OX and may be denoted by IK. Therefore

formula (4) may be written as

so that the problem of finding the moment of inertia may be

reduced to the solving of two problems of the type of the

first part of this section.

Ex, 3. Find the polar moment of inertia of an ellipse with respect to

the origin,

In Ex. 2 we found IK for a quadrant of the ellipse. For the entire

ellipse it is four times as great, since moments of inertia are added bydefinition. Hence

264 REPEATED INTEGEATION

By a similar calculation Iy-

Theiefoie ^0=4 ffa* (2+ &*)

If the area is leplaced by a plate of mass Af, this result gives

/ =\. M(a* + I")

If polar coordinates aie used, the element of area is rd6dand the distance of a point in an element from the origin is

Hence, in (2), dA = rdBdr and R = r. Therefore

r-(6

In practice it is usually convenient to integrate first wit

respect to r, holding 8 constant. This is, in fact, to find tr

polar moment of inertia of a sector with vertex at 0.

Ex.4. Find the polar moment of inertia of a cncle with lespect to

point on its circtimference

Let the circle be placed as in Fig 110 Its equation is then (Ex 1, 5]

r = 2 a cos 0, where a is the radius If we take any element rdQdr anfind I for all elements which lie in the

same sector with it, we have to add the

elements i*dQdr, with r ranging from to

t v where rt is the value of ? on the cncle

,

and therefore >l= 2 a cos We have

fVrffltft = i 7*(IB = 4 a4

/o1

We have finally to .rm these quantities,

with 1anging from to + - We have** *

I=J*la*cQ&6d6= I

~I TIG. 110

If M is the mass of a ciicular plate, this result, multiplied by p,

Ex. 5. Find the polar moment of inertia of a cucle with respect to i

centei.

Heie it will be convenient first to find the polar moment of inertia ofring (Fig 111). We integrate first with respect to 6, keeping r constanWe have

MOMENT OF INERTIA 265

which is the approximate area of the ring Swrdr multiplied by the

square of the distance of its inner cucumference from the center Wethen have, by the second integration, ^

= fa

Jo

If M is the mass of a circular plate, this

result, multiplied by p, givesI / / f\ S \ U \\ \

X

The moment oE inertia of a solid

of revolution about the axis of revo-

lution is the sum of the moments of

inertia of the circular sections about

the same axis ; that is, of the polar]FIG< m

moments of inertia of the circular sections about their centers.

If the axis of revolution is OY, the radius of any circular

section perpendicular to OF is x and its thickness is dy. Its

mass is therefore pirx*dy ; and therefore, by Ex. 5, its moment

of inertia about Y is\ pirtfdy. The total moment of inertia of

the solid is therefore

X\pTr \ x*dy.

Ex. 6. Find the moment of inertia of a circular cone about its axis.

Take the cone as in Ex. 4, 85. Then we have

But, if M is the mass of the cone, we have M =

Therefore / = fa Ma*.

EXERCISES

1. Find the moment of inertia of a rectangle of base & and alti-

tude a about a line through its center and parallel to its base.

2. Find the moment of inertia of a triangle of base b and altitude

a about a line through its vertex and parallel to its base.

3. Find the moment of inertia of a triangle of base & and alti-

tude a about its base.

4. Find the moment of inertia of an ellipse about its minor axip

and also about its major axis,

266 REPEATED INTEGRATION

5. Find the moment of inertia of a trapezoid about its lower base,

taking the lower base as b, the upper base as a, and the altitude as 7*.

6. Find the moment of inertia about its base of a parabolic seg-

ment of base b and altitude a.

7. Find the polar moment of inertia of a rectangle of base I and

altitude a about its center.

8. Find the polar moment of inertia about its center of a circular

ring, the outer radius being rzand the inner radius rr

9. Find the polar moment of inertia of a right triangle of sides

a and b about the vertex of the right angle.

10. Find the polar moment of inertia about the origin of the area

bounded by the hyperbola xy=6 and the straight line aj+y 7=0.

11. Find the polar moment of inertia about the origin o the area

bounded by the curves y=xz and y = 2 as2

.

12. Find the polar moment of inertia about the origin of the area

of one loop of the lemniscate r2 = 2a? cos 2 0.

13. Find the moment of inertia of a right circular cylinder of

height Ji,radius r, and mass M, about its axis.

14. Find the moment of inertia about its axis of a hollow right cir-

cular cylinder of mass M, its inner radius being rv its outer radius ra ,

and its height h.

15. Find the moment of inertia of a solid sphere about a diameter.

16. A ring is cut from a spherical shell whose inner and outer radii

are respectively 5 ft. and 6 ft, by two parallel planes on the same

side of the center and distant 1 ft and 3 ft respectively from the

center Find the moment of inertia of this ring about its axis

17. The radius of the upper base and the radius of the lower base

of the frustum of a right circular cone are respectively r^and r# and

its mass is M. Find its moment of inertia about its axis.

89. Moments of inertia about parallel axes. The finding of a

moment of inertia is often simplified by use of the followingtheorem ;

The moment of inertia of a body about an oasis is equal to its

moment of inertia about a parallel axis through its center of gravity

plus the product of the mass of the body ly the square of the

distance between the axes.

MOMENT OF INEKTIA 267

We shall prove this theorem only for a plane area, in the

two cases in winch the axes lie in the plane of the figure or

are perpendicular to that plane. We shall also consider the

mass of the area as equal to the vKT *

area, as in 88.

Case I. Wlien the axes lie in tJie

plane of the Jiyure.

Let the area be placed as in

Fig. 112, where the center of grav-

ity (^ #0 lf} taken as the origin

(0, 0) and where the axis of y is

taken parallel to the axis LK, Labout which we wish to find the

moment of inertia. Let x be the distance of an element dxdyfrom OF, and ^ its distance from LK. Then, if I

gis the

moment of inertia about OF, and Jzthe moment of inertia about

LK, we have ^^1=11 x*dxdy, It i I xldxdy. (1)

JJ JJ

Moreover, if a is the distance between OF and LK, we have

so that, by substituting from (2) in the second equation of (1),

we have rf* **. npJi- x*dxdy + 2 a I I xdxdy + a* I dxdy. (3)

Now, by 84, ndxdy^Ai by 85, \( xdxdy=Ax = 0, since

by hypothesis x = ; and, by (1), the first integral on the right

hand of (8) is Ig

. Therefore (8) can be written

7,-7,+ oU, (4)

which proves the theorem for this case.

Case II. When the axes are perpendicular to the plane of

the figure.

We have to do now with polar moments of inertia. Let the

area be placed as in Fig. 118, where the center of gravity is

268 REPEATED INTEGRATION

taken as the origin, and P is any point about which we wish

the polar moment of inertia. Let Igbe the polar moment of

inertia about 0, and Ip

the polar moment of inertia about P.

Draw through P axes PX' and ,

PY1

parallel to the axes of coor-

dinates OX and OY. Let Ix and

Itlbe the moments of inertia

about OX and OY respectively,

and let 1^ and Jtf/ be the moments

of inertia about PX' and PY'.

Then, by (5), 88,/ /

JJ.--

Moreover, if (a, 6) are the coordinates of P, we have, byCase I,

1^=1^ a*A, /,=/ + VA. (6)

Therefore, from (5), we have

/,==/,+ <V+&V> (7)

which proves the theorem for this case also.

The student may easily prove that the theorem is true also

for the moment of inertia of any solid of revolution about an

axis parallel to the axis of revolution of the solid.

Ex. Find the polar moment of inertia of a circle with respect to a pointon the circumfeience.

The center of gravity of a circle is at its center, and the distance of anypoint on its circumference from its center is a. By Ex 5, 88, the polarmoment of a circle about its center is |ira*. Therefore, by the above

theorem,I,= i* + o (F) t *.

This result agrees with Ex 4, 88, where the required moment of inertia

was found directly

EXERCISES

1. Find the moment of inertia of a circle about a tangent.

2. Find the polar moment of inertia about an outer corner of

a picture frame bounded by two rectangles, the outer one being of

dimensions 8 ft. by 12 ft, and the inner one of dimensions 5 ft. by 9 ft.

SPACE COORDINATES 269

3. Find the moment of inertia about one of its outer edges of a

carpenter's square of which the outer edges are 15 in. and the inner

edges 13 in

4. Find the polar moment of inertia about the outer corner of

the carpenter's square in Ex. 3.

5. From a square of side 20 a circular hole of radius 5 is cut,

the center of the circle being at the center of the square. Find the

moment of inertia of the resulting figure about a side of the square.

6. Find the polar moment of inertia about a corner of the square

of the figure in Ex. 5.

7. Find the moment of inertia of a hollow cylindrical column

of outer radius raand inner radius 1\

about an element of the inner

cylinder.

8. Find the moment of inertia of the hollow column of Ex. 7

about an element of the outer cylinder.

9. Find the moment of inertia of a circular ring of inner radius ^and outer radius ?-

aabout a tangent to the outer circle.

10. A circle of radius a has cut from it a circle of radius75 tangent

i

to the larger circle Find the moment of inertia of the remaining

figure about the line through the centers of the two circles

11. Find the moment of inertia of the figure in Ex 10 about a

line through the center of the larger circle perpendicular to the line

of centers of the two circles and in the plane of the circles.

90. Space coordinates. In the preceding pages we have be-

come familiar with two methods of fixing the position of a

point in a plane ; namely, by Cartesian coordinates (x, y), and

by polar coordinates (r, 0). If, now, any plane has been thus

supplied with a coordinate system, and, starting from a point

in that plane, we measure another distance, called 0, at right

angles to the plane, we can reach any point in space. The quan-

tity will be considered positive if measured in one direction,

and negative if measured in the other. We have, accordingly,

two systems of space coordinates.

1. Cartesian coordinates. We take any plane, as -3T0F, in which

are already drawn a pair of coordinate axes, OX and OF, at

right angles with each other, Perpendicular to this plane at

2TO BEPEATED INTEOKATION

z

the origin we draw a third axis OZ (Fig. 114). If P is any

point of space, we draw PM parallel to OZ, meeting the planeXOY at Jf, and from M draw a line par-

allel to OF, meeting OX at L. Then for

the point P (#, y, z), OL x, LM y,

and MP = z. It is to be noticed that

the three axes determine three planes,

XOY, YOZ, and ZOX, called the coor-

dinate planes, and that we may just as

readily draw the line from P perpendic-

ular to either the plane YOZ or ZOX and

then complete the construction as above.

These possibilities are shown in Fig. 115, where it is seen that

x = OL = NM= SE = TP, with similar sets of values for y and s.

2. Cylindrical coordinates. Let XOYbe any plane in which a fixed pointis the origin of a system of polar coor-

dinates, and OX is the initial line of that

system (Fig. 116). Let OZ be an axis

perpendicular to the plane XOY at 0.

If P is any point in space, we drawfrom P a straight line parallel to OZuntil it meets the plane XOY at M. yThen, if the polar coordinates of M in

the plane XOY are r = OM, 6 = XOM, and we denote the dis-

tance MP by s, the cylindrical coordinates of P are (r, 0, g).It is evident that the axes OX and

OZ determine a fixed plane, and that

the angle 6 is the plane angle of the

dihedral angle between that fixed planeand the plane through OZ and the

point P. If SP is drawn in the latter

plane perpendicular to OZ, it is evident

that OM= SP = r and OS = MP = z,

The coordinate r, therefore, measuresthe distance of the point P from the axis OZ, and the coordinate2 measures the distance of P from the plane

Z

116

SUEFACES 271

If the line OX of the cylindrical coordinates is the same as

the axis OX of the Cartesian coordinates, and the axis OZ is the

same in both systems, it is evident, from 51, that

a; = r cos 0, y == r sin 0, = z. (1)

These are formulas by winch we may pass from one systemto the other. It is convenient to notice especially that

(2)

91. Certain surfaces. A single equation between the coordi-

nates of a point in space represents a surface. We shall give

examples of the equations of certain surfaces which are impor-tant in applications. In this connection it should be noticed

that when we speak of the equation of a sphere we mean the

equation of a spherical surface, and when we speak of the

volume of a sphere wo mean the volume of the solid bounded

by a spherical surface. The word sphere, then, indicates a sur-

face or a solid, according to the context. Similarly, the word

cone is used to denote either a conical surface indefinite in

extent or a solid bounded by a conical surface and a plane

base. It is in the former sense that we speak of the equationof a cone, and in the latter sense

that we speak of the volume of

a cone. In the same way the

word cylinder may denote either

a cylindrical surface or a solid

bounded by a cylindrical surface

and two piano bases. This double

use of these words makes no con-

fusion in practice, as the context

always indicates the proper mean-

ing in any particular case.

1. Sphere will center at origin. Con-

siderany sphere (Fig. 117)with its center

at the origin of coordinates and its radius

equal to a. Let P be any point on the surface of the sphere. Pass a plane

through P and OZ, draw PS perpendicular to OZ, and connect and P.

Then, using cylindrical co6rdinates, in the right triangle OPS, OS=z, SP^r,and OP as a, Therefore . 8s= a9. (1)

FIG. 117

272 REPEATED INTEGRATION

This equation is satisfied by the cyhnducal coordinates of any.point onthe surface of the sphere and by those of no other point Jt is therefore

the equation of the sphere in cylindrical

coordinates.

By means of (2), 90, equation (1)

*"***+? + *,* (>)

which is the equation of the sphere in

Cartesian coordinates.

2. Sphere tangent at origin to a cooi 1-

dinate plane. Consider a sphere tangentto the plane XOY at (Fig 118). Let

P be any point on the surface of the

sphere Let A be the point in which the

axis OZ again meets the sphere Pass a

plane through P and OZ,connect A. and

P, and P,and diawPS perpendicular

to OZ Then, using cylindrical coordi-

nates, OS = z, SP = i, and OA = 2 a,

where a is the radius of the sphereNow OAP is a right triangle, since it is inscribed in a semicircle,and PS is the perpendicular from the vertex of the right angle to the

hypotenuse. Therefore, by elementary plane geometry,

~SP*=OS' SA = OS(OA - 08)

Substituting the proper values, we have

X

FIG. 118

Z

(3)

which is the equation, of the sphere in cylindricalcoordinates.

BJ (2) 90> equation (3) becomes

#a+fya +22 2 a? = 0, (4)

which is the equation of the sphere in Cartesian

coordinates.

3 Right circular cone. Consider any right circular

cone with its vertex at the origin and its axis alongOZ (Fig 119). Let a be the angle which each elementof the cone makes with OZ. Take P any point on the surface of the cone,pass a plane through P and OZ, and draw PS perpendicular to OZ. Then

SPSP = r and OS = z. But = tan SOP - tan a Therefore we haveC/o

r = gtano (5\\VJ

as the equation of the cone in cylindrical coordinates,

FIG. 119

SURFACES 273

By 2, 90, equation (.">)becomes

0, (6)

as the equation of the cone in Cartesian coordinates,

As explained above, we have heie used the word cone in the sense of a

conical surface If the cone is a solid with its altitude h and the radius of

its base a, then tan a - In this case equation %li

(5) or (6) is that of the curved surface of the

cone only.

4. Surface of revolution. Consider any sur-

face of revolution with OZ the axis of revolution

(Fig 120). Take P any point on tho smface

and pass a plane through P and OZ In the

piano POZ draw OR peipendicular to OZ and,

fiom P, a straight line perpendicular to OZmeeting OZ in S If we regard OR and OZ as

a pair of i octangular axes foi the plane POZ,tho equation of the cuive CD in which the

plane POZ cuts the surface is FIG 120

=/(r) f (7)

exactly as y ~f(x) is the equation of a curve in 12

But CD is the same curve in all sections of the surface through OZTherefore equation (7) is true for all points P and is tho equation of the sur-

face in cylindrical coordinates. Whan the plane POZ coincides with the

planeXOZtr is equal to x, and equation (7) becomes, z

for that section, z=.ffy.\ > /-g\

Hence we have the following theorem :

The equation of a surface of revolution formed by

revolving about OZ any curve in the plane XOZ maybe found in cylindrical coordinates by writing rfor x

in the equation of the curve.

Tho equation of the surface in Cartesian coor-

dinates may thtn be found by placing r = Vr* + y*.

For example, the equation of the surface formed

by revolving the parabola2 =4o: about OZ as

an axis is za as 4 ? in cylindrical co&rdinates, or

s*sBlO(.efl + y

8) in Cartesian coordinates.

5. Cylinder. Consider first a right circular cylinder with its axis along

OZ (Fig. 121). From any point P of the surface of the cylinder draw 7J S

perpendicular to OZ. Then 8P is always equal to , the radius of the

cylinder. Therefore, for all points on the surface,

reo, (9)

121

274 REPEATED INTEGRATION

which is the equation of the cylinder in cylindrical cobrdinates. Reduced

to Cartesian coordinates equation (9) becomes

x*+y*=a?, (10)

the equation of the cylinder in Cartesian coordinates.

More generally, any equation in x and y only, or in r and 6 only, repre-sents a cylinder. In fact, either of these equations, if interpreted in tho

plane XOY, represents a curve, but if a line is drawn from any point in

this curve perpendicular to the plane XOY, and P is any point on tins lino,

the coordinates of P also satisfy the equation, since z is not involved in

the equation. As examples, the equation ya =4:x xepresputs a parabolic

cylinder, and the equation r = a sin 3 & represents a cylinder whoso base is

a rose of three leaves (Fig. 65, p. 144).6 Ellipsoid Consider the surface defined by the equation

c2 (11)

If we place z = 0, we get the points on the surface which Ho in tho XO I'

plane These points satisfy the

equation .*

- + *= = ! (12) _.^

and therefore form an ellipse.

Similarly, the points in theZOX plane lie on the ellipse

S + 73= 1> (13)

and those in the YOZ plane lie

on the ellipse

6s c2 (14)

FIG. 122

The construction of these

ellipses gives a general idea ofthe shape of the surface (Fig 122). To make this more precise, let UHplace * =

jin

(11), where zx is a fixed value. We have

which can be writtenc2

==1, (16)

SURFACES 275

As long as z < c3, equation (16) represents an ellipse with semiaxes

a v'l"

and b A/1 ^ By taking a sufficient number of these sections

we may construct the ellipsoid with as much exactness as desired.

If z = c8 in (16), the axes of the ellipse reduce to zero, and we have a

point IfSja > ra, the axes are imaginaiy, and there is no section.

1 Elliptic paraboloid. Consider the surface

where we shall assume, foi defimteness, that c is positive.

If we place z = 0, we get 2

-2+ f8 = >^ ftA /ift

(17)

(18)

which is satisfied in real quantities only by x and y = 0. Therefore the

XOY plane simply touches the surface at

the origin

If we place z = c, we get the ellipse

1' (19)

which lies in the plane c units distant from

the XOY plane

If we place y = 0, we get the parabola

and if we place a? = 0, we get the parabola

FIG. 123

The sections (10), (20), and (21) determine the general outline of

surface. For more detail we place z = sx and find the ellipse

c c

so that all sections parallel to the XOY plane and above it are ellipses

(Fig. 123).8. Elliptic cone. Consider the surface

,

I3-!! 0. (23)

Proceeding as in 7, we find that the section z = is simply the origin

and that the section, z = cis the ellipse

276 REPEATED INTEGRATION

If we place x 0, we get the two straight lines

i =-t*-

and if we place y = 0, we get the two stiaight lines

a = -zc

(25)

(20)

The sections we have found suggest a cone with an elliptic baso. To

prove that the surface really is a cone, we change equation (SJiJ)to cylin-

cotirdmates, obtaining

^ + E^\ ra = ~- nm2 /i2 / /(2 V <*

B I/ / C

Now if 6 is held constant in (27), the coefficient of rz is constant, and thi

equation may be written T1 J r = kz, (28)

which is the equation of two straight hues in the plane through OK

Z Z

FIG 124

determined by & = const Hence any plane trough OZ cuts the surfacetwo straight lines, and the surface is a cone (Fig. 124).

9. Plane Consider the surface

n

Ax + ED + Cz -f D = 0. /29\The section z = is the straight hne ffff (Plg 125) with the equation

Ax + Vy + DssQ, (80)the secfcon y = is the straight line LH with the equationAt + a+DszO,

(31)and the section , = ia the straight hne L/C with the equation

By + Cz + T) = 0.

VOLUME 277

The two lines (31) and (32) intersect OZ in the point L (O, 0,J

,

unless C = 0. Assuming for the present that C is not zero, we change

equation (29) to cyhndncal cooidmates, obtaining

(.4 cos 6 + B sin d)r + Cz + D = 0. (33)

This is the equation of a stiaight line LN in the plane B = const It

passes through the point L, which has the cyhndncal coordinates r = 0,

2 _ ~i; and it meets the line KH, since when z = 0, equation (33) is the

same as equation (30). Hence the sui face is covei ed by straight lines which

pass through L and meet KH. The locus of such lines is clearly a plane

We have assumed that C in (29) is not zero. If C = 0, equation (29) is

Q. (34)

The point L does not exist, since the lines corresponding to HL and KLare now parallel. But, by 5, equation (34) lepresents a plane parallel to

OZ intersecting XOYm the line whose equation is (34) Theiefoie we

have the following theorem

Any equation of the first degree represents a plane.

92. Volume. Starting from any point (#, y, z) in space, we

may draw linos of length dx, dy, and dz in directions parallel

to OX, OF, and OZ respectively, and on these lines as edges

construct a rectangular parallelepiped.The volume of this fig-

ure we call the element of volume dV and have

(1)

For cylindrical coordinates we construct an element of volume

whose base is rd6dr( 84), the element of plane area in polar

coordinates, and whose altitude is dz. This figure has for its

volume dV the product of its base by its altitude, and we have

d7=rd0drdz. (2)

The two elements of volume dV given in (1) and (2) are

not equal to each other, since they refer to differently shaped

figures. Each is to be used in its appropriate place. To find

the volume of any solid we divide it into elements of one of

these types. ,

To do this in Cartesian coordinates, note that the a-coordinate

of any point will determine a plane parallel to the plane YOZ

278 REPEATED INTEGRATION

and x units from it, and that similar planes correspond to the

values of y and & We may, accordingly, divide any ruquirotl

volume into elements of volume as follows:

Pass planes through the volume parallel to Y0% ami <ta*

units apart. The result is to divide the required volume into

slices of thickness dx, one of which #is shown in Fig. 126. Secondly,

pass planes through the volume

parallel to JTO^and dy units apart,

with the result that each slice is

divided into columns of cross sec-

tion dxdy. One such column is

shown in Fig. 126.

Finally, pass planes through the

required volume parallel to XQYand dz units apart, with the result

that each column is divided into

rectangular parallelepipeds of dimensions das, dy, and fe, Oneof these is shown in Fig. 126.

It is to be noted that the order followed in the aboveexplanation is not fixed and that, in fact, the choice of be-ginning with either a or y or 4 and the subsequent orderdepend upon the particular volume zconsidered.

A similar construction may bemade for cylindrical coordinates.In this case the coordinate &determines a plane through OZ.We

accordingly divide the volumeby means of planes through OZmaking the angle d0 with eachother. The result is a set of slicesone of which is shown in Fig. 127 Fro. 127

VOLUME 279

Finally, these columns are divided into elements of volume

by planes parallel to XOY at a distance dz apart. One such

element is shown in Fig. 127.

When the volume has been divided in either of these ways,it is evident that some of the elements will extend outside the

boundary surfaces of the solid. The sum of all the elements

that are either completely or partially in the volume will be

approximately the volume of the solid, and this approximationbecomes better as the size of each element becomes smaller.

In fact, the volume is the limit of the sum of the elements.

The determination of this limit involves in principle three in-

tegrations, and we write

=I | |

dxdyds (3)

or V= CCCrd6drdz. (4)

In carrying out the integrations we may, in some cases, find

it convenient first to hold z and dz constant. We shall then

be taking the limit of the sum of the elements which lie in a

plane parallel to the XOY plane. We may indicate this by

writing (3) or (4) in the form

V=* Cdz CCdxdy or V= Cdz CCrdOdr. (5)

But, by 84, 1 1 dxdy = A and \\rd6dr~A, where A is

the area of the plane section at a distance from XO Y. Hence

(5) is rF= / Adss, (6)in agreement with 26.

**

Hence, whenever it is possible to find A by elementary means

without integration, the use of (6) is preferable. This is illus-

trated in Ex. 1.

In some cases, however, this method of evaluation, is not

convenient, and it is necessary to carry out three integrations.

This is illustrated in Ex. 2.

280 REPEATED INTEGRATION

Ex. 1. Fmd the volume of the ellipsoid + =1.

By 6, 91, the section made by a plane parallel to XOY is an ellipse

with semiaxes a %/! ^ and Z> %/! - ^ Therefore, by Ex. 1, 77, its area

/ za\c

is waft (1 1 Hence we use formula (6) and have\ c2/

r el sa\ 4

V = irab I 1 1 -]ds=:- irdbc.J_ c \ c2/ 3

Ex. 2. Find the volume bounded above by the sphere a:2-f ?/

a + s2 = 5 and

below by the paraboloid a;2 + y

z = 4 z (Fig 128).

As these are stu faces of i evolution, this example may be solved bythe method of Ex. 1, but in so doing we need two integrations one

foi the sphere and the other foi the paraboloid We shall solve the

example, however, by the other method in order to illustrate that method

We fii st reduce our equations to cyhn- ^dncal coordinates, obtaining lespectively

r* + z* = 5 (1)

and r1 = 4 z (2)

The surfaces intersect when r has the

same value in both equations; that is,

whenz* + 4 z = 5, (3)

which gives z=loiz = 5 The latter

value is impossible ,but when z = 1, we

have r = 2 in both equations Theiefore

the surfaces intersect in a circle o radius 2 in the plane z = 1

lies duecHly above the circle r = 2 in the XOY plane.

We now imagine the element tdQdrdz inside the surface and, holding

r, 0, dd, dt constant, we take the sum of all the elements obtained by varyingz inside the volume These elements obviously extend from 2 = s

lin the

lower boundary to z = z2 in the upper boundary. From (2), z1= and,

from (1), 22 V5 r8. The first integration is therefore

This circle

rdddr MOdr.

We must now allow 6 and r so to vary as to cover the entire circle ? == 2

above which the required volume stands.' If we hold Q constant, r varies from to 2. The second integration is

therefore

VOLUME 281

Finally, 9 must vary from to 2 IT, and the third integration is

/5V5

If we put together what we have done, we have

/>2ir /.a ftF=f/ /Jo Jo t/,

EXERCISES

1. Find the volume bounded by the paraboloid = y? + ^ and

the planes x = 0, y = 0, and g = 4.

& C^ 9*2

2. Prove that the volume bounded by the surface - =-5 4-

fa

and the plane * = c is one half the product of the area of the base

by the altitude.

3. Find the volume bounded by the plane = and the cylinders

SBB-f f = 2 and y* = a* az.

4. Find the volume cut from the sphere r2 4-2= a2

by the cylinder

r = a cos 6.

5. Find the volume bounded below by the paraboloid r*= a and

above by the sphere r2 4-2 2 a# =

6. Find the volume bounded by the plane XOY, the cylinder

a-a {> 2/a_ 2 ax = 0, and the right circular cone having its vertex at 0,

its axis coincident with OZ, and its vertical angle equal to 90

7. Find the volume bounded by the surfaces ra= &, = 0, and

y sa a COS 0,

8. Find the volume bounded by a sphere of radius a and a right

circular cone, the axis of the cone coinciding with a diameter of the

sphere, the vertex being at an end of the diameter, and the vertical

angle of the cone being 90 .

9. Find the volume of the sphere of radius a and with its center

at the origin of coordinates, included in the cylinder having for its

base one loop of the curve i* ;= a2 cos 2 6.

10, Find the volume of the paraboloid a? 4- f = 2 * cut off by the

plane #~x 4-1.

11. Find the volume of the solid bounded by the paraboloid1 and the plane * = *.

dm

282 EEPEATED INTEGRATION

93. Center of gravity of a solid. The center of gravity of a solid

has three coordinates, x, % z, which are defined by the equations

I xdm I ydm I

*= J'

' v =Jr

' -I dm I dm I

where dm is the mass of one of the elements into which the

solid may be divided, and #, y, and a are the coordinates of the

point at which the element dm may be regarded as concentrated.

The derivation of these formulas is the same as that in 85

and is left to the student.

When dm is expressed in terms oi space coordinates, the

integrals become triple integrals, and the limits 'of integration

are to be substituted so as to include the whole solid.

We place dm = pdF, where p is the density. If p is constant,

it may be placed outside the integral signs and canceled from

numerators and denominators. Formulas (1) then become

7x= CxdV, Vy=CydV, Vz*= CzdV. (2)

Ex. Find the center of gravity of a body bounded by one nappe of a

right circular cone of vertical angle 2 a and a sphere of radius a, the center

of the sphere being at the vertex of the cone.

If the center of the sphere is taken as the origin of coordinates and the

axis of the cone as the axis of 2, it is evident from the symmetry of the

solid that x = y = Q. To find z we shall use cylindrical coordinates,

the equations of the sphere and the cone being respectively

r2 + z2 =s oa and r z tan a.

As in Ex. 2, 92, the surfaces intersect in the circle r a sin a in

the plane z = a cos a. Therefore

/>2r a a tin a />Va8 r8

V=\ I / rdOdrdz = S Tra8 (1- cos a)JO /o t/rotnar

o \ ./

n nZv na sin a /-\/a2 r*

and \zdV-\ \ \ rzdOdrdz = Iwa4 sina cr.t/ i/O /0 t/rctnrt

w

Therefore, from(2), 5 s? f a (1 + cos a).

CENTER OF GRAVITY 283

EXERCISES

1. Find the center of gravity of a solid bounded by the paraboloidy 7 /

- = '

-f 7; and the plane = cc a2 1r

L

2. A ring is cut from a spherical shell, the inner radius and the

outer radius of which are respectively 4 ft. and 5 ft., by two parallel

planes on the same side of the center of the shell and distant 1 ft

and 3 ft. respectively from the center. Find the center of gravity of

this ring.

3. Find the center of gravity of a solid in the form of the frustum

of a right circular cone the height of which is h, and the radius

of tho upper base and the radius of the lower base of which are

respectively r^ and ?'2

.

4. Find the center of gravity of that portion of the solid of

Ex. 2, p. 73, which is above the plane determined by OA and

OB (Fig. 31).

5. Find the center of gravity of a body in the form of an octantQn Aa i*

ej*

of the ellipsoid~ + 73 + -= = 1.e a? o* <?

6. Find the center of gravity of a solid, bounded below by the

paraboloid az r* and above by the right circular cone * + f = 2 a.

7. Find the center of gravity of a solid bounded below by the

cone * SB r and above by the sphere r* + #2= 1.

8. Find the center of gravity of a solid bounded by the surfaces

JB = 0, i* + *a ** i3,and r =

94. Moment of inertia of a solid. If a solid body is divided

into elements of volume c?F, then, as in 88, the moment of

inertia of the solid about any axis is

1= CtfpdV** p C&dT, (1)

where JB is the distance of any point of the element from the

axis, and p is the density of the solid, which we have assumed

to be constant and therefore have been able to take out of the

integral sign. If M is the total mass of the solid, p may be

determine^ from the formula JtfwpK

284 REPEATED INTEGRATION

If the moment of inertia about OZ, which we shall call 1^ is

required, then hi cylindrical coordinates JR = r and dV= r$6drdz,

so "that (1) becomes

Iz = p CCCr*d0 drds. (2)

If we use Cartesian coordinates to determine/,,,

we have

and dV= dxdydz, so that

'xdydz. (3j

Similarly, if Iyand Ix are the moments of inertia about OY

and OX respectively, we havef\ f\ SI

(4)

In evaluating (2) it is sometimes convenient to integrate

with respect to z last. We indicate this by the formula

Ia = p Cdz CCr9 dO dr. (5)

But I I i^dddr is, by 88, the polar moment of inertia of a

plane section perpendicular to OZ about the point in which OZintersects the plane section. Consequently, if this polar momentis known, the evaluation of (5) reduces to a single integration.This has already been illustrated in the case of solids of revolution.

A similar result is obtained by considering (3). In fact, the

ease with which a moment of inertia is found depends upon a

proper choice of Cartesian or cylindrical coordinates and, after

that choice has been made, upon the order in which the integra-tions are carried out.

Equation (3) may be written in the form

>WCy*dxdydt, (6)

and the order of integration in the two integrals need not bethe same. Similar forms are derived from (4).The theorem of . 89 holds for solids. This is easily proved

by the same methods used in that section.

MOMENT OF INERTIA 285

Ex. Find the moment of inertia about OZ of a cylindrical solid of

altitude h whose base is one loop of the curve r a sin 3 6,

The base of this cylinder is shown in Fig 65, p. 144. We have, from

formula (2),-

In8fl

wheie the limits are obtained as follows:

First, holding r, 6, d6, dr constant, we allow z to vary from the lower

base s = to the nppei base z = li, and integrate. The result phra dOdi is

the moment of inertia of a column such as is shown in Fig 127. Wenext hold and d& constant and allow r to vary from its value at the

origin to its value on the curve r = a sin 3 0, and integrate. The result

\ /7i4 ain43 Qd& is the moment of inertia of a slice as shown in Fig. 127.

Finally, we sum all those slices while allowing to vary from its smallest

value to its largest value ^ The result is zz

The volume of the cylinder may be computed from the formula

a"1*- /idBinSO nh

V =I f i rdBdrdz = AJo Jo Jo

Therefore Af= -faphePir and Js =JI/ 2.

EXERCISES

1. Find the moment of inertia of a rectangular parallelepiped

about an axis through its center parallel to one of its edges.

2. Find the moment of inertia about OZ of a solid bounded bythe surface t= 2 and $ = r.

3. Find the moment of inertia of a right circular cone of radius a

and height h about any diameter of its base as an axis.

4. Find the moment of inertia aboxit OZ of a solid bounded by2 Q

the paraboloid*

=* ~ + ^ and the plane = e.

5. Find the moment of inertia of a right circular cone of height li

and radius a about an axis perpendicular to the axis of the cone at

its vertex.

6. Find the moment of inertia of a right circular cylinder of

height h and radius a, about a diameter of its base.

7. Find the moment of inertia about OZ of the portion of the

sphere ?* 4- *" ~ aa out out by the plane and the cylinder

r ** a cos 6.

286 REPEATED INTEGRATION

8. Find the moment of inertia about OX. of a solid 'bounded by

the paraboloid r* and the plane = 2.

9. Find the moment of inertia about its axis of a right elliptic

cylinder of height h}the major and the minor axis of its base being

respectively 2 a and 2 b.

10. Find the moment of inertia about OZ of the ellipsoid

t+t + t =iI

'

7,2' 2^

GENERAL EXERCISESn n n

1. Find the center of gravity of the arc of the curve x* -I- y* = *,

which is above the axis of x.

2. A wire is bent into a curve of the form 9y2 = a?

8. Find the

center of gravity of the portion of the wire between the points for

which x = and x = 5 respectively.

3. Find the center of gravity of the area bounded by the curve

ay*= a? and any double ordmate.

4. Find the center of gravity of the area bounded by the axis

of x, the axis of y, and the curvej/

2 = 8 2 x>

5. Find the center of gravity of the area bounded by the curves

y = x9 and y = -5 > the axis of x, and the line x = 2

6. Find the center of gravity of the area bounded by the axesof x and y and the curve x = a cos8

^, y = a sins <

7. Find the center of gravity of the area bounded by the ellipsejg2

&

~z + a =1 (a > *) tlie <Hfle as2 + y

2 = a,and the axis of y.

8. Find the center of gravity of the area bounded by the parabolav? = 8 y and the circle cc

a + f= 128

9. Find the center of gravity of the area bounded by the curves**~ a (y

-&)= 0, a2

ay = 0, the axis of y, and the line = o,

^

10. Find the center of gravity of an area in the form of a semi-circle of radius a surmounted by an equilateral triangle having oneof its sides coinciding with the diameter of the semicircle.

11. Find the center of gravity of an area in the form of a rec-

tangle of dimensions a and I surmounted by an equilateral triangleone side of which coincides with one side of the rectangle which isb units long.

GENERAL EXERCISES 287

12. Find the center of gravity of the segment of a circle of

radius a cut off by a straight line b units from the center.

13. From a rectangle b units long and a units broad a semicircle

of diameter a units long is cut, the diameter of the semicircle

coinciding with a side of the rectangle. Find the center of gravityof the portion of the rectangle left.

14. Find the center of gravity of a plate in the form of one half of

a circular ring the inner and the outer radii of which are respectively'

xand >

15. In the result of Ex. 14, place rz=

r^ + AT- and find the limit as

Ar vO, thus obtaining the center of gravity of a semicircumference.

16. Find the center of gravity of a plate in the form of a T-square10 in across the top and 12 in. tall, the width of the upright andthat of the top being each 2 in.

17. From a plate in the form of a regular hexagon 5 in. on a side,

one of the six equilateral triangles into which it may be divided is

removed. Find the center of gravity of the portion left.

18. Find the center of gravity of a plate, in the form of the ellipser

o/~ + 75= 1 (a > i),

in which there is a circular hole of radius c,

the center of the hole being on the major axis of the ellipse at a

distance d from its center.

19. Find the center of gravity of the solid formed by revolving!K2

?/2

about OY the surface bounded by the hyperbola -j j$=1 and the

lines?/= and y = &.

a

20. Find the center of gravity of the solid generated by revolvingabout the line % = a the area bounded by that line, the axis of a;, andthe parabola y

a = hx.

21. Find the center of gravity of the segment cut from a sphere of

radius a by two parallel planes distant respectively \ and hz (hz > 7^)

from the center of the sphere.

22. Find the moment of inertia of a plane triangle of altitude a andbase b about an axis passing through its center of gravity parallel to

the base.

23. Find the moment of inertia of a parallelogram of altitude aand base b about its base as an axis,

288 BEPEATED INTEGRATION

24. Find the moment of inertia of a plane circular ring, tho inner

radius and the outer radius of which are respectively 3 m and 5 in.,

about a diameter of the ring as an axis.

25. A square plate 10 in. on a side has a square hole 5 in. on a

side cut in it, the center of the hole being at the center of tho jilafco

and its sides parallel to the sides of the plate. Find the moimmt of

inertia of the plate about a line through its center parallel to ouo

side as an axis.

26. Find the moment of inertia of the plate of Ex. 25 nbout one of

the outer sides as an axis.

27. Find the moment of inertia of the plate of Ex. 20 about one

side of the hole as an axis.

28. Find the moment of inertia of the plate of Ex 2tf about ono

of its diagonals as an axis

29. A square plate 8 in. on a side has a circular hole 4 in. in

diameter cut in it, the center of the hole coinciding with tho cuntor

of the square Find the moment of inertia of the plate about a lino

passing through its center parallel to one side as an axis.

30. Find the moment of inertia of the plate of Ex. 29 about a

diagonal of the square as an axis

31. Find the moment of inertia of a semicircle about a tangent

parallel to its diameter as an axis.

32. Find the polar moment of inertia of the plate of Ex. 25 aboutits center.

33. Find the polar moment of inertia of the entire area bounded

by the curve T-2 = a? sm 3 6 about the pole.

34. Find the polar moment of inertia of the area bounded by thocardioid r = a ( 1 + cos

ff)about the pole.

35. Find the polar moment of inertia of that area of tho circler = a which is not included in the curve r a, sin 2 6 about thopole.

36. Find the moment of inertia about OF of a solid bounded bythe surface generated by revolving about OY the area bounded by thecurve </ x, the axis of y, and the line y 2.

37. A solid is in the form of a hemispherical shell the innerradius and the outer radius of which are respectively and r , Findits moment of inertia about any diameter of the base of the shell asan axis.

GENERAL EXERCISES 289

38. A solid is in the form of a spherical cone cut from a sphereof radius a, the vertical angle of the cone being 90. Find its

moment of inertia about its axis.

39. A solid is cut from a hemisphere of radius 5 in by a rightcircular cylinder of radius 3 in, the axis of the cylinder being

perpendicular to the base of the hemisphere at its center Find its

moment of inertia about the axis of the cylinder as an axis.

40. An anchor ring of massM is bounded by the surface generated

by revolving a circle of radius a about an axis in its plane distant

b(b > <i)from its center. Find the moment of inertia of this anchor

ring about its axis.oj2 ?/"

41. Find the moment of inertia of the elliptic cylinder + 77 =1a* (r

(a > ft),its height being h, about the major axis of its base.

42. Find the center of gravity of the solid bounded by the cylinder= 2 a cos 0, the cone ss = r, and the plane x =43. Find the moment of inertia about OZ of the solid of Ex. 42.

44. Find the volume of the cylinder having for its base one loopof the curve r = 2 a cos 2 0, between the cone = 2 r and the plane=46. Find the center of gravity of the solid of Ex. 44.

46. Find the moment of inertia about OZ of the solid of Ex. 44.

47. Find the volume of the cylinder having for its base one loopof the curve r ss a cos 2 & and bounded by the planes = andft =s x + 2 a.

48. Find the moment of inertia about OZ of the solid of Ex. 47

49. Find the volume of the cylinder r= 2 a cos included between

the planes * = and * =s 2 x + a.

60. Find the moment of inertia about OZ of the solid of Ex. 49.

61. Through a spherical shell of which, the inner radius and the

outer radius are respectively rland r

a,a circular hole of radius

a (a < fj)is bored, the axis of the hole coinciding with a diameter

of the shell. Find the moment of inertia of the ring thus formed

about the axis of the hole.

ANSWERS

[The answers to some problems are intentionally omitted.]

CHAPTER I

Page 4 ( 2)

1. 21^. 4. 100ft per second 7. 2J2. l^jj. 5. 88 07 mi. per hour for entire trip. 8. l^mi per hour.

3. 40J\>. 6. 1,26. 9. 08.4.

Page 5( 2)

10. 106 8.

Page 7 ( 3)

1. 06 ft. per second. 2. 128 ft. per second.

Page 8 ( 3)

8. 128 ft. per second. 5. 68 ft per second.

4. 74 ft pci bccoud. 6. 52 ft. per second.

Page 11 ( 5)

1, 12 ii ;24

j.8. 85

;82

;6. 5. 5, 4, when t = 2

; 10, 6, when t = 3

2. 16 j 14. 4, 42557. 8. 8aia + 2M + c ; 6at + 26.

Page 13 ( 6)

1. i- sci. ft. per second. 2, 0. 3. 87rra cu. in. per second.TT 2?r

4, 4wra. 5, 167irsq. in. per second.

Page 14 ( 6)

6. 8irr. 7. 8 (edge)3

. 8, birr*. 9. 18. 10. 2ir.

CHAPTER II

Page 18 ( 7)

3. 4fcB -2. 4. :- 5. 2 ;

8, -,_S

.7. tf + B+1. 8, 8-~

201

292 ANSWERS

Page 21 ( 9)

1. Increasing if x > 2; decreasing if x < 2.

2. Increasing if x > g , decreasing if x < |.

3. Increasing if x < ; decreasing if x > .

4. Increasing if a < , decreasing if sc > .

5. Increasing if aj < 2 or x > 1, decreasing if 2 < K < 1.

6. Increasing if x < 5 or x > S, decreasing if 5 < x < 8.

7. Increasing if < 1 or a: > ; decreasing if 1 < x < fj.

8. Always increasing.

9. Increasing iftc< lor ^<a<l; decreasing if 1 < x < J or a; > 1.

10. Increasing if x > 1, decreasing if x < 1.

Page 24 ( 10)

1. When t < - 1 or t > I, when !<<!.

2. When 5;when t > 5.

3. When t < 2 or t > 4; when 2 < i < 4.

4. Always moves in dnection in winch a is measxirod.

5. When t > f ,when < $.

6. Always increasing.

7. Always decreasing8. Increasing when t > 2

, decreasing when t < 2.

9. Increasing when t > , decreasing when i < $10. Increasing when < < f , decreasing when < > |.

Page 26 ( 11)

1. TrA8 . 2. 6 irh sq. ft. per second.

Page 27 ( 11)

8. 2 cm. per second. 5. 0.20 TT

4. 20 9 sq in. per second. 6. 64 cu. ft per second.Tl 3

8(t tal holffht>*'

8. 4 TT (2 + 12 ( + 36) ,t is thickness.

Page 31 ( 13)

1. 1.46. 3 0.46; 2.05. 5. 2.41.

2. - 2 07. 4. 1.12; S 93. G. 2.52.

Page 35 ( 14)

1. 8a;-y_9 = 0. 6, + y + l = *12. tan-ij.2. 2x + Sy + 3 = 7. aj + 2y + 8 = 0. 18, tan-*12.

3. 21aj-2y-13 = 0. 8. 4x- Sy-1 = 0. 14.*.

4-J/ + S =

,_-M.-4y-.6-0. W. i.-Sy-M-o.

6. V3a;-y-2V3-2=! 0. 10. 5a: - Qy - 4 = 0. W, (-Itfr, 2ft)*

*The symbol tan- 1jj- represents the angle whose tangent is \ (cl. 46).

ANSWEES 293

Page 39 ( 15)

1. (- I >|).

2, (2, 4J).

3. (0, 4), (2, 0).

4, (1, 7), (3, 3).

6. (- 2, 0), (1,-

0).

6. (-1, - 3), (3,29).

Page 43 ( 17)

1. aSsq.in.

2. Length is twice breadth.

Page 44 (5 17)

6, Depth is one-half side of base.

7, 2 portions 4 ft. long ;4 portions 1 ft. long

-o jjt.8, Breadth =

9. (- 3, 10), (1, 2)

10. 278 + 27^-86 =11. 18a5

18. tan-ijj

3. 5ft.

4. 50.

- - 28 = 0.

5.

, ..

; depth =

9, Altitude = ^-f- 5base - ? (P - perimeter)

4 4

10, 2000 cu. in.;2547 cu. in

1L Height of rectangle = radius of semicircle ,semicircle of radius,

512. -7= in.

VS

Page 46 ( 18)

1, 426ft.

Page 47 ( 18)

8. 40ft. 5. 676ft.

7. y = 2x8 + xa -4x+6.3, 05ft.

4, 88$ ft.

Page 49 ( 19)

1. 7$. 2. 1

Page 53 ( 20)

8. 0.0001 ;0.000001 ; 0.00000001

9. 0.000009001 ;0.000000090001.

3, 62J. 4. 36. 5.

Page 54

1. 72 sq. in.

8,

16

Sircu. in.

6.

10. 000003 sq. in.

11, 456 58 ou. in.

4. 27.0054 ou. in.

5. 28.2749 cu m.

6. 606 0912.

7. 0012.

8. 5.99934

294 ANSWERS

Page 55 (General Exercises)

_'

o 2a A 4 xg _ j t

13. Increasing if x > 2,

'

(a-

x)2

" '

(x2 + I)

2' '

2Vajs

'

decreasing if < - 2.

14. increasing if f<o;<ora;>2, decreasing if a; < f or<j< x < 2.

15. Increasing if x > -, decreasing if y, < - .

2 o, 2(t

16. Increasing if x < ^= or x >~ , decreasing if ^= < x < =V3 V3 V3 V3

4/7 4 ft

17. Increase if x < -,decrease if x > -

3 o

Page 56 (General Exercises)

18. 1< t < 5 19. 2 < * < 5,4.

20. Up when < i < 6J ,down when 6 < < 12.

21. Increasing when t > 4, decreasing when t < 4.

22. D increasing when < 3, v deci easing when t > 3, speed increasing when

2<i<3ori>4, speed decreasing when !S<2or3<i<423. Increasing when !<<2ori>3, decreasing when *<lor2<i<324. 0055 in. per minute

25. 8 6 in. per second 27. x + 2 y + 6 = 0.

26. 1 sq in per minute. 28. 7x+

Page 57 (General Exercises)

29. x- 2 = 31. 2x-y + 3 = 0. 83. (-, 8|).

80. x - 2y - 7 = 0. 32. tan-l 34. (lj, 0)

35. (2,-

2). 41. 10.36. (- 1, 13), (5,

-95). 42. tan-i TV

37. (- 3, 13), (1,-

19). 43. x - y - 11 =38. (- 4, 20). 44. (1,

-1), (- J,

-

Page 58 (General Exercises)

46. 6f ft. long 52. y - x2 + Bx - IB

47. Altitude of cone is | radius of sphere. 53. y = x8 x* + 7 x

54.85^.i** a ^ *u48. Altitude=

;sideof base=^_ M 2g

49. 2 pieces 3 in long, 3 pieces 1 in. long 56. 20150 600ft. 57. 72.

51. 56ft 59. 0.0003.

Page 59 (General Exercises)

60. 0.00629. 64. 0.09 cu in 67. 24.0024 sq. in.

62. 288 TT cu in. 65. 0.0003. 69, 0.4698.

63, 161.16 cu. in, 66. 854.1028j 353.8972,

Page 66( 23)

1. 8J 3. 52 T S .

2. 23,} 4. 166fr

Page 67 ( 24)

1. 160ft

Page 68( 24)

4. When m

Page 70( 25)

1. 8$ T.

Page 71 ( 25)

6. Approx. 2418 Ib.

6.

2. 140ft.

; 83ft.

3. 57Jft.

5, 8000irft.-lb.

2. 2JT. 3. 3T.

Page 75 ( 26)

1. 21 TT.

625V8~~

7. 585}T.

11. 2 1 ft. from upper side

8. 84A *.

4. l^w.5. 6577r.

cu in.

7. 8

8. 2|

Page 76 (General Exercises)

1. 6Jft. 5. 20.

2. 81ft.

3. 10 ft"'~3~

4. 8JJ mi- ' 8.

Page 77 (General Exercises)

13. Twice as groat. 17.

14, Jft T.

16. 16 to. 35

16. 68^TT 19. 96 IT.

Page 78 (General Exercises)

24. STT. 26. 115J. 26.

29. 728,049 ft.-lb.

Page 81 ( 28)

1. &a + i/8 - 8as -f

CHAPTER IV

11 cs 0.

0.

6, 8a~

9. 234| T10. 2 T.

9. 25f IT

10. 213JTT.11. 38|

8.

9. JfT12. Reduced to | original

pressure

20. 34lcu.m.

21.^-128. (aft

2-Jft")ir

27. 9 28. 204|80. 5301 ft Jb.

3. (-3, 5); 5.

*.(- *.!);

296 ANSWERS

Page 84 ( 30)

M-2,o). MO, -if) 7. as/, ft

a- (o.i). 5. 8H ft.. iom/5

MH,0). 6. lOVlOft.8.

im

Page 85 ( 30)

9. y2 + 6a;-9 = 0. 10. x*-4x-12y + 16= 0.

Page 87 ( 31)

2- (0, 3), (0, V6),

5. 9x2 + 25^a - 36x - 189 =6. 49a;2 + 24 y2 _ 120y _ 144 _ .

Page 91 ( 32)

I- (3,0), (

4. o,f

2- (2,0), (Vl3, 0), 3a;2y=0; 13. (0, V2), (0, V5),4. (2V,0); (4,0),

Page 102 ( 36)

1. 18s2 + 22a;_3

8).9-

_'

~(a;2 -9)2

'

11. -.

2112.

13.

7 2a! _i + ^*2 ^'

14 .8.

'

ANSWERS 297

15 -" . -. 18.'

3i tr -i>) \' (x ~"a) (JR

-6) V(aa + x2

)8

ii.M 1 jr811 *"'

. 19.,

22.

v'W-.tfi (x + l)Vxa -l V(x2

+9)8

'r ~ ]. 20.

v*--* vV + 8)* V(l + x)*

Page 104 ( 37)

<tt/ jea Vi/ + x Vy x __x

2 2a8x

* ""'U, 1/& ai

a/// g !_. 8> "i'

'

'* *"--J i ^ ,,'S* ""ftw* 8u8

'

xa 2X2

e'

-3 !

Page 105 (38)Sa O. 2, x~7y + 5 = 0. 3. (-2.-1). 4.

PAgel06(88)-. 10, tan-18. 12.

8.*". . 5 5ton" 1 A- 11. tan-i|. 13.

"i *

Page 110 (40)1. x-r 8 5

V44.9tK ,. foa Bin

)

a, x^d/-!)'4 ,V4tf+i. \ 20 20

3. x*-0x + = 0j SVll7-48t + l6ia . B a sill2a

4. (8,1).9 '

'4*

11. yaa

\ "aj

8Vi Moond. 2. 12.6 ft, per second. 3, ^u, perminute.

4, Circle;!5-fl, per secona (x- distance of point from *dl).

x

, 2,64 ft. per second.

Pftge 113 (41)perminute. 8. 6.6 ft. per second.

6. 0.18 cm. por second. ?, 0.21 in. per mumi*.

, ^ti^t.per 8ccond, where, is the distance of top of ladder, and IB

^dtonoe of foot of ladder, fromDase of pyramid.

298 ANSWERS

Page 114 (General Exercises)

20. 3

21. x

8

27. tan-

' ^ "

28. -; tan- 17.2i

30. tan-i .

5

35.

81. -, tan- 1 --

2' 2

~ 7TOn i n 1132. -,lan

33. x$ + ifi

= 2 ato - <W8?'

OJ2

-1- 4'

(i2 + I)

2

Page 115 (General Exercises)

37. 20 ft per second,

lOVEft per second, (100,20).

1a

' va

39. Velocity in path = VCKC + a5

40.

Page 116 (General Exercises)

46. 08 ft. per minute

47. 01 in. per minute.

48. |f sq in per minute

49. 04 in. per second.

50. Length is twice breadth

51 Other sides equal.

Page 117 (General Exercises)

57. 2 64 in 58.

41. 6 8 mi per houi,28 8 mi

42.

43.

t

4|250-3i

a

_

'

3 \ 04- 13

ft pei second, whore3s

Vs2 - 400

s is length of lope from man

to boat.

44. 06 ft per minute

45. sJu ft per minute.

52. Breadth, 9 m, depth, 9V3 in.

53. Length = | bieadth

54. Side of base, 10 ft,

depth, 5 ft

55. Depth = ^ side of base.

56. Radius, 3 in., height, 3 in.

1

Vi'60. 8 mi from point on bank nearest to A.

bm , , bn ,

a mi. on land,

mi in water

59.

61. 41|}

mi tiavel ou land

Vn2 m2

63. l^f-hr.

Page 118 (General Exercises)

Vn2 - mr

64. v/lOO mi per hour.

Page 126 ( 44)1. 15 cos 5 x

2. sec2-.

3. 2sm22zcos2x.4. 5 sin 10 x.

65. Velocity m still water = mi per hour

66. Base = aVI ,altitude = | b

CHAPTER V

6. 5 sin2 5 x cos8 5 aj.

_ o60!, &X7. 5 sec2 tan .

2 2

8. -8csc8 3a;ctn3x.

ANSWERS 299

Page 127 ( 44)

9. hin-1

.

10.

11.

12. 2 sec x (sou x + tana-)2

.

Page 129 ( 45)

. S_. Mil---

3. 7T, 5.

4. At moan point of motion ;at

extreme points of motion.

2

13. 2 cos 4 x.

14. 9tan4 3aj.

15 . 2 sec 2 x (sec2 2 x + tan2 2 x)

16 sm8 2 a; cos2 2 a;

17. - ?}

18. -?.

5. At extieme points of motion, at

moan point of motion.

6. 2V(fc-

3) (5-

s) ; 4(4-8)7. TT.

8. 10, 2?r.

18.

Page 136 ( 48)

1. 1)3.= 9.42 ft. per second ; % = =F 86.40 ft. per second.

Page 137 ( 48)

3. 8 radians per unit of time.

Page 138 ( 49)

a b cos <t)

'

Page 141 ( 50)17V17

2. 63

6.

300 ANSWERS

Page 145 ( 51)

17. Origin ,

(V3, j).18. Origin ,

( a^j|, |). 19. Origin ,

(2, |

21. ^8^20 =22. r

23. r

24. r

25. x - a =26. x2 + V* - 2 ax =27.

28.

Page

Page

9.

10.

11.

18.

19.

148 ( 52)

149 (General Exercises)

2(l + sec2 2x).

sec4 (3x + 2).

cos2 (2- 3 x).

sec2 (x y) + sec2 (x + y)

sec2 (x y) sec2 (x + y)

1 .x

1. IT tan-* 1 3. 0.

21.

20. -

tan4 2 a;.

l_

(x + 1) Vx-1

V2 + x - x2

27.

2(x + l)Vx

1

xV49x2 -l

4

ig t

5 5

14 8csc24x(ctn4x + l)

15. a tanox sec2 ox.

16. 8 cos8 2 x sin 4 a; cos 6 x.

1

. x)Vx3 2x

2

V3-4x-4x2

2~3xVo x2

24.

25.

26.

Page

29.

30.

31.

(x2 -l)Vx2-2

150 (General Exercises)

fcVa2 sm2^ + 62 cos2^.

2^_V41,

5 = 3; 2

35.

j47. Origin ;

/

^""*"

O*

/ A.

34. 2aVi.48. Origin; (^,tan-i

2).

49. Origin;

Page 151 (General Exercises)

51. -=COS(?

52.

53. (x8 + y

2)2 + 2 aa;(x

a + y2)- a

54, tan-i.s= 0.

ANSWERS 301

65. 0) tan-12 57. -. 59 IflVift

56. 0, :,ton-i8V5. Kg*6 ' 72 '

2 68'4 61. V2ft

62. At an angle tan- 1/*; with ground.

Page 152 (General Exercises)

63.12m. 65. a. 68. 15 sq ft,10.04 sq. ft. per second.

64. sVBft. 67. 0.1 ft per second. 69. 26 7 mi. per minute.

nn /i /i ,

&2 8in0COS(? \ . . ,.., ,70. / b sin & + , )

times angular velocity of AS. where = angle\ Vaa -6a sin2 0/

CL4.B.

71,<*~ 2)

2

+ fa- 3

)2 - 1

;V9 sin2 * + 4 cosat

,where I = (2 k + 1) |

78. 5^-^ = 1; OsecSiVtan2 3< + 4sec2 3<.

Page 153 (General Exercises)

73. sin 20. 75. tan-^Vi74. aVl+cos2

o5, fastest when 75. tan- 1^

x = far; most slowly when 77. 0, tan- 1 3 Vs.

a5 = (2A! + l)-. 78. tan-i^, tan-i4V22 79. tan-i3; tari-ij.

CHAPTER VI

Page 162 ( 55)

(The student is not expected to obtain exactly these answers , they are given

merely to indicate approximately the solution.)

1. y =0.62 x- 0.70. 2. 1 = 0.00172)

Page 163 ( 55)

3. ^=s0.80(2.7)^. 4. c=0.010(0,84). 5. a=0.0000000048Z3'o. 6 j" = 10.

Page 165 ( 56)

1 1 12. e- 2ac(3cos3a; 2sin8x).

" ~*< n j. 113, Ctll"" 1

!!/.

14.

3. 2 a; a**-Una."'

Vi^"+~4* 16,

4. a n^ _.,

9 16.e" + e-

5|rf j. A -. i

'

10. - 4 sec 2*. W- 2 see's.

20:4.8 n 2(e8a! e-2a?

) ig. J; .

- - - "' '

eaa+e-is*'

asVx + l

302

Page 167 ( 57)

ANSWERS

2. ?/= 45.22eoola:. 4.

Page 168 ( 57)

5. P = 10000 e 0229* 6. c = 0.01e- n44B<. 7, 2 mm,

Page 168 (General Exercises)

*10. p = 0.018 1 + 24. *11. Load = 102 - 6 6 length.

Page 169 (General Exercises)

Page 171 (General Exercises)

31. 2"\/2e~ 2i,2e~ 2 * 33.

87.(l + 1)*

35. Vic*.

x'

238.

Page 176 ( 59)

CHAPTER VII

3.8-2.

, ,a;8

. 2s5,

17x73. a; + H---h

8 15^

315

2i 41 61

4- ^"4" "" -L. - ____ - ____ __ JL2 32-4 6

+2 . 4 6 7

e.

V2

-2 2 2s 33

T5 7

T '

X~2l~3! + "

j

* Statement in regard to answeis to exeicises in 55 is true of tins answer,

9. 0.0872

10, 4695,

ANSWERS 303

Page 178 ( 60)

5 2 52 8 58

___ + .___ _____ + .

8/ ir\8

1 V V ,

oy 2 21 2 si

6 E +ito+

2 2! 2 31

8. 0.7193 9, 0.8480 10. 8.0042

Page 179 (General Exercises)

_ +'

5~

2*

5'

2 1 2_

'

8 I

a j.^ - 1 ^!_I^ ?!.1

z %~x

a '21 2*31

10. 1 - 2cc + 2 2 - 2fc8 + . . 14. 9659.

11 i j.2aj8 4a* 15. 0.61CO.u . a H. x-- + ...

T21 8! 5!^

'

18. 0.69815; 10986.

13 3.4.?! + ! + ^!+...,19> -22314; 1.6094.

Page 180 (General Exercises)

88, 2.0805.2e

a;8 a6

_

a7

M, 2.9625.' "*"

3 5(2 I)

""

7(8 !)

304 ANSWERS

CHAPTER VIII

Page 183 ( 61)

8j/2

, .

xy x2'

xy y35.

y X

6. ---,(K + y)* x + y

'

s._L_, *_. 7i *

x2 +2/2'

x2 +i/2y

'

tf

y x 14. , > , s. , ;

VI xay2 Vl x2^2 v x2 + ya

Page 185 ( 62)

x2 y*1.

-5 5-2. e" sm (x y) 3,

Page 187 ( 63)

1. 000061 2. 0012. 3, 2^

Page 188 ( 63)

4.0.018m 5.00105 6.0015m. 7. 6320ft. 8.0.0064.

Page 191 ( 64)

1. -2. 2. -| 4. - 5. - J . o. 6. 0;0Vx2 + y2

Page 192 (General Exercises)

8. - 14.33 cu ft. 10. 0.5655 sq. in.; 11. 3.0 in 13. 2.206 sq. in. per second9l 1735 05756sq.m 12. 035m, 15. 4.4 uq. in. per second.

Page 193 (General Exercises)

16. _iyJ5. 17 /y-tan-ia fit ia J V5 .

5a tan A

, &*. is. -cos a .

'Sina, 1.

CHAPTER IXPage 198 ( 66)

. -p if* -p uiiiij,. 7. i(x* + 4)a. 14 1 , .

2. g\(3x+7)x4 8. llnW-im.'

~aln

(1 + c ^).9. ^In(2x + sin2x) 14. ~Jcos*2x.

10. - 1 W-

4. inz __! L. 3(x-smx) 16a: 2.758 .1

1)]. a 17. _12. ln(x

8 + Sx2 + 1). ig,i. 20.

ANSWERS 805

Page 202 ( 67)

, 1 ,3x1. -sin- 1 -

3 4

i.sin-

1

Vs

4. sec-1xV8.i a

5.

'

V7 V?

V21

4x-33

6x-511. -^-sm-i

Vs 5

V3

*5

-=-.VlO

12.

13.

V5

V2 V2

16.? 17. ~.4

Page 204 ( 68)

3.

5.

. Jln(8a+Vo"-l).

-p

n

-r-

15 5sm- 1--2V4-x2.

ft

18. v. 19. ~- 2C36

11.

In (3

1 . 2x-620

U2x + 5'

7.

8.

9.

-

2VlC 8X+V15--7:

2 Vi5

~

w, 4 to-..

Page 207 ( 69)

1, - ^ COB (S 05-2).2, ~Jsin(4~2x).3, Jsec(8x 1).

4, 4tan?.4

2 8x

6. $ln sin 6 as,

7.

3a; + 5

12. ^Lln-*"- 7-V5 2 x - 8 + V5

'

V33 2 x + 5 -t- V33_1 . 4x-l-Vl3

14. =rln _2 VlS 4x-l + Vl3

8 + V5

15.

16,

18. -4=1]V2

19. ln;

20. iln

8. 2 esc -

9. ^ln [sec (4 x + 2) + tan (4 a + 2)].

10. ctn(8-2a).11. In (esc x etnas) + 2 cosx.

13.

14, a cosx.

306 ANSWERS

_

4. Vx2 1 tan- 1 V.e2 1

8. -

5.

x ,x6 sin- 1 --

V4-X2 2

9. -4Vx2 -4

10. |.11.. 12.- 18. ^(9\/8-10V5). 14.. 15. 2

Page 216 ( 74)

1. (8 x 1) e3* 7. J (2 cos + sin a) e2 *.

8. J(x2 + 2xsincc + 2cosx).

9. 4-2Ve10. J (81 In 8 -96).11. i(-2).12. l(7T-2).

2.

3. zcos-ix Vl x2

4. xtan- 1 3a-j^ln(l+ 9a2

5. la^Beo-^x6. sin z (In sin a; 1)

Page 217 ( 75)

4. InX2 -l

2)2

2.

Page 220 (General Exercises)

--.a; 2 as

2

5. In'

ANSWERS 307

Page 221 (General Exercises)

T./, (2 + CST

)S

. 9. -\/8 a; + xa

8.ij (1 + 2s + a;*)*. 10. za + a: + In (a

-1).

11. n\, [B fdu (2 z - 1)- 3 sm (2

11) ] .

12. - un - /3 COB* - + 4 cos2 - + B\.8 f>\ 5 5 /

13. a1

,)ctn 4 x

(]5 + 10 ctn2 4 x + 3 ctn4 4 a) .

14. ,V, t8 HucB(-c - 2)

- 5 suc(a--

2)].

15. i tau(jt-

1) + In tan (.r-

1)

16. Ju (7 c\scfi 2 x 5 c'sc7 2 tr) .

11 i(MoSa--7)v/8ee8j.

18. Vcac2x19. ~?% (8 ctn 5 x + 8 ctn3 5 a:)

Vctn 6 x

20. T jt,j(lB3 34sina 4o; + Own4 4 a) \/wn4a5.

21. -7'

fl (9 ctn 5 x + 4 ctn 6 a1

)\/ctn8 5 x.

23, sin- 3

V5

1 . 2 sr + 8- sec- 1 -!

5

24.^sin-i-^-.s aVa

,. JL O 3525 - - i ppf**" J-

i , *'

V6 V6

4 2

21 i Bec-i^ti.2 2

28. J-sec-i^li.'sVs Vs

52;r-5

,2J!-1

OB35.

Vl630. sin- 1

31.

00 i ,23; + 833. = sin- 1 =

V2 V8

36. _Vo

38

39.

Vn

Vc

Vn

2V6 V5

Page 222 (General Exercises)

41.

-ft 1 .a -2

42 - Rl Tl **" 1 -i"T rrm-i

2 2

1 1-2

43. -i-soc-i^-.aVo V61 9-S. - JL nv

** "^T SGC "*"* *

48. ln(oj-|-Vxa -7).

49.

45, 4 1

46.1

VlO

51.

62. i Inj^-h Valfi + 7). _53, 2 V(ca + 4 + In (a; 4- V a + 4).

54. V8ica+l --

4r1V21

55, -i= In 8 + -Ox).

308 ANSWERS

157. =ln(5x + 2 + V25x2 + 20 x 5)

VH

58. ln(3x-4 + V9x2 - 24x + 14)

4 V6 x V& - 2i ^ ^

62.

59. -6V? 8x+V7

^in^?- 2.

4 Vo x V5 + 2

2V21

63. ln-.6 x

.. 1, 2x-564. -In

5 x

.^2V6

70. l

__ 1 , x-1bo. in

12 3x +

4V6 2x-

67.15

5" x + 2

Page 223 (General Exercises)

753 3 .a; x 9 . 2a;-x -sm8 -cos- sin .

8 4 3 3 16 3

68. ta

71. (tan3x ctnSx).72. In [sec(x-g) + tan

(aj-

)]

73. cos2x.

74. ln(secx + tanx).

Gx-3)

91.

(In 5)8

' 106.

ANSWERS 809

Page 224 (General Exercises)

107. V5-2 112. 6? 118.

V5'

113. 7. 119 .

108'0114.^11X2.

109>c2 ~ e-

1 9 + 4V2 WO-

110.~^j-

.

'

^^/gn

14'

121. 2 - In 3.

, I / ! \ 116. STT 122.Til ... ........ V A" i t ft*/'

lu 5'

117. UvS. 123.

CHAI?TEB X

Page 229 ( 77)

10. 2w2a8.

2 ?H^V2 A/8^13> T T 16<

11. Jjfira8

'

~8" 14. 259jw. 16. JTTW.

Page 232 ( 79)

1. 2a2. T 2 5

aa V2 7, HTT.^

2 7m_2

'JJ," 3 8.J. 10. 4-(7T-2).

4n' 4. ~~. 6. |7ra2 9. 407T.

*

2

Page 234 ( 80)

1. ^L.;

ft is radius of semicircle. 2.;a is xadius of semicircle.

2 2u 3

IT 7T0

Page 235 ( 80)

6. 7ra8 . 7. 100 revolutions per minute. 8. 5.64 Ib. per square inch,

Page 236 ( 81)

8), 4. Ca. 7.

I 5. STr'a.

310 ANSWERS

Page 238 ( 82)

lm vka 5 686 1 ft.-lb

'a '12 6. 2kca2,k is the con-

2, 22f ft.-lb 4. 196,350 ft.-lb stant ratio

7. - mi.-lb ; B is radius of earth in miles 8. 2 irCR + a

Page 239 ( 82)

9. 1.76ft.-lb., 1 56ft.-lb.

Page 239 (General Exercises)

1 Ssm-i| 3. 16 -12 In 3 6. 1^ , 15T\~

,

5__. 9V3247T + - .

5 15

10 ^TTCE2 _

11. ^ Trfc2Vi^jj , fet

and kz are the values for k in the equation j/

2 =

12. -.. 13.15

Page 240 (General Exercises)

14. dfrrf. 15.4FGn4-l) 16 17.

19.

5

.,15 15 '

23. _ 4

Page 241 (General Exercises)24. 400irlb. 28. 123 T 31. 21 J.

.

26. 441b. n 8

34. -(8ir+9V3) 35.'

J.D '

Page 242 (General Exercises)

861-4 37.

^(8+7T), (57T-8), (8-7T)

88'

T* ^ * 41.950 42. -(In 9-1).

ANSWERS 311

48,|

45. -""2 8a3

Page 243 (General Exercises)

q T.

49. ~. 50. 50,000 ft.-lb. 51, 438,1 ft.-lb.a a*

CHAPTER XIPage 245 ( 83)

1.80 -In 8. 2. In 3. 8. 2Jf. 4. IT 1

Page 246 ( 83)

6. 14.' T" '

4"~

' '

~6~"

10, ~ (22-

TT). 11, Va (IT V2 - 4). 12. .

y 30

Page 254( 85)

(2cz (i (f^ 4- 4 c" 1^\ 4 a 'N/S~~r> . /0 . ) 6. On axis of qxxadrant, from

e+l 4e(e8--l) / ? . .

' Sv'

.

v 7 center of circle.

4. lira,~r-\ 7. Intersection ot medians.

8. /-, -\.5, On axis,

~ from vertex. \ 2 8 /

10. On axis, distant4 a* + 3

f+ 0fca

fr0m L!!'

16. On axis, distant (radius) from base. ir. On axis, distant - from base.a

16. On axis, distant - from base. 18. Middle point of axis.

Page 256( 86)

3, On line of centers, distant(ri

*"*,? from center of circle of radius r,.

rx2 + r|

4, On axis of shell, distant ^ a~~

*' from common base of spherical surfaces.

5, Middle point of axis.8 (

r *" ^ )

Page 257 ( 86)

6, On axis,a 1

- distant from base,

7, On axis, $ of distance from vertex to base.

8, (4$, 4|), the outer edges of the square being; taken as OJTand OF.

312 ANSWERS

9. On axis, distant 4 8 from corner of square

10. On axis, distant 3.98 m from center of cylinder in direction of larger ball.

11. On axis, distant 8.4 ft. from base of pedestal.

Page 260 ( 87)

3. base x altitude

4. ^,a = altitude and 26 = base of segment.

15

5-7rg

; a = altitude and 26 = base of segment.5

6. 27ra26, 8ira6.

7. ^(b + Sc), 7r[2ca + 26c + 62 + (6 + 2c) Va2 + 62]

Page 268 ( 89)

1 McP, a = radius.

Page 269 ( 89)

9

10.

11.

4) 632a8

9

7Sffffl4

326

8. Tra8 9. (37T + 20 16V2). 10. u,

ANSWERS 313

Page 283 ( 93)

2t On axis of ring, distant 2ft. from / Q a*

center of shell. 6- f^i -^V

3. On ax1S,distant

7i(r

' + 2 Vt + '.") / 6 + 8

4W + V. + T/)?.

(0,0,

+

from upper base

, , 8/00 8 2(2b2 - 2

) \4. On axis, from base. Ot

(

u u.

('

8 \ 8[b8 -(b2 -aa

)&]/

Page 285 ( 94)

1. ^5 Jlf (a2 + 62

), where Jf is mass, and a and & are the lengths of the sides

perpendicular to the axis.

2. Atf M.8, jf M(3 a2 -1- 2 W) 5. ^M(a + 4 /i) 2 a2 (15 TT - 26)

4. J Jfcf (a8 + 62

).6. ^M(8 a

2 + 47i).

'

25 (3 TT - 4)

Page 286 ( 94)

14.9T8. i^..3

Page 286(General Exercises)

.

8. .. 9. }flf(a2 + &8). 10.

3

. f , oV a; =\ 7 /

6 is the ordhiale. 852

10. On axis, distant _ from base of triangle and away from

.

semicircle.

11, On axis, distant4a*+ 2ab-^ + &*

from base.

Page 287 (General Exercises)

18, On axis of segment, distant ^ T.IL1'" ^ from center

of circle,8(7m8

314 ANSWERS

18. On axis, distant, ,

~~from center of semicircle.

8(86 -Tra).

! J- 7* f J- f^\14. On axis of plate, distant

* v ' a 1J from center of circles.

gp V 2""" I/

15. On axis, distant 1 from center of circle7T

16. On axis of square, 8 in. florn bottom.

17. On axis, distant - in. from center of hexagono

c2d18. On axis, distant from center of ellipse

ab ~ C96

19. On axis of solid, distant from smaller base16

20. On axis, distant from base ^ of distance to top.

21. On axis of segment, distantl~

*2~"

ll '-" fmrn ofmtnr4-T^n^fh Ji ~\ f] 8 'ii **"* \jouuv3i

of sphere21 2

22. T^lfa2

23,

Page 288 (General Exercises)

24. *j-2l. a*

25. JJL5 u 31> S7 (15 ^ - 32)J 34.16

26. *fM. a 1S radius.

27. &&M 35.

32. %JiM. 30

29 XPJli_4 486'

30, uyi_4Tr]33. ~. 37t

Page 289 (General Exercises)

38-3Q-18-5V2;.

42. (^,0,39. J/i&M 43.

40.

1 1 ly/OLO . y in, 44

48 '

32+

3465' 49 ' 3ira8 80.

51. ^ __

^13838103

INDEX

(The numbeis refer to the pages)

Abscissa, 28

Acceleration, 9, 21, 186

Algebraic functions, 70

Amplitude, 128

Angle, between curve and radius

vector, 140

between curves, 104

between straight lines, 35

vectonal, 142

Angular velocity and acceleration, 185

Anti-sine, 180

Approximations, 53, 187

Arc, differential of, 106, 146

Archimedes, spual of, 145

Area, as double integral, 246

of ellipse, 225

of plane curve, 47, 225

in polar coordinates, 280

by stimulation, 60

of suiface of revolution, 259

Asymptote, of any curve, 80, 02

of hyperbola, 90

Average value See Mean value

Axis, of cooidmates, 28

of ellipse, 86

of hyperbola, 00

of parabola, 82

Cardioicl, 145

Cartesian equation, 100

Cartesian space coordinates, 260

Catenary, 157

Center of_ gravity, of any solid, 282

of circular arc, 258'

of composite area, 255

of half a parabolic segment, 25f

of plane area, 251,

Center of gravity, of plane curve,

250

of quarter circumference, 250

of right circular cone, 253

of sextant of circle, 252

of solid of revolution, 252

Circle, 79, 148

of cuivature, 140

Circular measure, 119

Cissoid, 98

Compound-interest law, 166

Cone, circular, 272

elliptic, 275

Constant of integration, 45, 194

Coordinates, 27

cylindrical, 270

polar, 142

space, 269

Curvatuie, 189

Curves, 91

Cycloid, 137

Cylinder, 278

Cylindrical codrdinates, 270

Definite integral, 62, 194

Derivative, 15

higher, 40

partial, 181

second, 39

sign of, 20, 40

Differential, 50

of arc, 106, -146

of area, 64

total, 185

Differential coefficient, 51

Differentiation, 15

of algebraic functions, 94

815

316 INDEX

Differentiation, of exponential and

logarithmic functions, 163

of implicit functions, 102

of inverse trigonometric functions,

131

partial, 181

of polynomial, 18

of trigonometric functions, 124

Directrix of parabola, 81

Distance between two points, 79

Double integration, 244

e, the number, 155

Eccentricity, of ellipse, 87

of hyperbola, 90

Element of integration, 64

Ellipse, 85

area of, 225

Ellipsoid, 274

volume of, 280

Elliptic cone, 275

Elliptic paraboloid, 275

Equation of a curve, 29

Equations, empirical, 159

parametric, 109

roots of, 30

Equilateral hyperbola, 90, 92

Exponential functions, 154

Palling body, 6, 8

Focus, of ellipse, 85

of hyperbola, 87

of parabola, 81

Force, 128

Formulas of differentiation, 101, 124,

131, 163

of integration, 195, 199, 202, 205,

207, 217

Fractions, partial, 216

Functions, 15

algebraic, 79

exponential, 154

implicit 102

inverse trigonometric, 130

logarithmic, 154

trigonometric, 119

Graphs, 27

of exponential functions, 157

of inverse tiigonometnc functions,130

of logarithmic functions, 157

in polar coordinates, 142

of trigonometric functions, 121

Hyperbola, 87

Implicit functions, 102

Increment, 16

Indefinite integral, 63, 194

Infinite integrand, 229

Infinite limits, 229

Integral, 45, 194

definite, 62, 194

double, 244

indefinite, 68, 194

Integrals, table oJ, 217

Integrand, 194

Integration, 45, 194

collected formulas, 217

constant of, 45, 194

by partial fractions, 216

by parts, 212

of a polynomial, 45

repeated, 244

by substitution, 208

Inverse sine, 130

Inverse trigonometric functions, 130

Lemniscate, 144

Length of curve, 235

Limit, 1

of

of (1 + h)\ 156

theorems on, 93

Limits of definite integral, 63

Line, straight, 81

Linear velocity, 135

Logarithm, 154

Napierian, 156

INDEX 317

Logarithm, natural, 156

Logarithmic spual, 168

Maclaurm's series, 173

Maxima and minima, 41

Mean value, 233

Measure, circular, 119

Moment o inertia, 200

of circle, 204

polar, 202

ol quadrant of ellipse, 262

of rectangle, 201

of solid, 283

of solid of revolution, 265

Moments of inertia about parallel axes,

260

Motion, in a curve, 107

simple harmonic, 127

Napierian logarithm, 166

Ordinate, 28

Origin, 27, 142

Pappus, theorems of, 259

Parabola, 81, 146

Parabolic segment, 83

Paraboloid, 275

Parallel lines, 38

Parameter, 109

Parametric representation, 109

Partial differentiation, 181

Partial fractions, 216

Parts, integration by, 212

Period, 128

Perpendicular linos, 84

Plane, 276

Polar coordinates, 142

Polar moment of inertia, 202

Pole, 142

Polynomial, derivative of, 18

integral of, 46

Power series, 172

Pressure, 68

theorem on, 267

Projectile, 110

Radian, 119

Eadms of curvature, 139

Eadius vector, 142

Bate of change, 11, 189

Revolution, solid of, 73

surface of, 259, 273

Hoots of an equation, 30

Rose of three leaves, 144

Second derivative, 89

sign of, 40

Segment, parabolic, 83

Senes, 172

Maclaurm's, 173

power, 172

Taylor's, 177

Sign of derivative, 20, 40

Simple harmonic motion, 127

Slope, of curve, 86

of straight line, 31

Solid of revolution, 73

Space coordinates, 269

Speed, average, 3

true, 6

Sphere, 271, 272

Spiral, logarithmic, 158

of Archimedes, 145

Straight line, 81

Substitution, integration by, 208

Summation, 66

Surfaces, 271

of revolution, 273

Table of integrals, 217

Tangent line, 88, 104

Taylor's series, 177

Total differential, 185

Trigonometric functions, 119

Trochoid, 138

Turning-point, 37

Value, mean, 238

Vector, radius, 142

Vectonal angle, 142

Velocity, 21, 107

angular, 135

318

Velocities, related, 111

Vertex, of ellipse, 86

o hyperbola, 90

ot paiabola, 82

of parabolic segment, 84

INDEX

Volume, of any solid, 277

of solid with parallel bases, 71

of solid of revolution, 73

Work, 237